Mappings and applications

Conformal maps, dispersion relations and pole expansions.

In this chapter 4

Tick a unit to add it to your PDF basket.

Conformal maps and applications

Conformal maps and applications

Method

Conformal mappings

Adapted

Add to PDF

An analytic map with non-vanishing derivative turns and rescales a small figure without distorting it, Mobius maps carry circles and lines to circles and lines, and a harmonic function composed with such a map stays harmonic — which is how an awkward two-dimensional Laplace problem is transported to a geometry one can solve by inspection.

Prerequisites: Analyticity and the Cauchy-Riemann equations, Complex numbers and elementary functions

NoteOriginal

What it means. Analytic functions are rigid in a very useful way. Near a point where $f'(z_0) \neq 0$ , the map $z \mapsto f(z)$ does to a small figure exactly what multiplication by the single complex number $f'(z_0)$ does: it turns it through $\arg f'(z_0)$ and scales it by $\lvert f'(z_0)\rvert$ . Every direction is turned by the same angle and every length scaled by the same factor, so angles between curves survive and small shapes are reproduced, merely rotated and resized. That is all "conformal" means.

Two consequences make it a working tool rather than a curiosity. First, a harmonic function pulled back through an analytic map is still harmonic, and a level set is still a level set, so a solved Laplace problem in one region becomes a solved Laplace problem in every region analytically equivalent to it. Second, the simplest non-trivial analytic maps — the Mobius maps $(az+b)/(cz+d)$ — move circles, lines, half-planes and disks around freely, and the power and exponential maps open and close corners. Between them they reach most of the two-dimensional geometries that electrostatics and ideal fluid flow actually present.

Definition(Conformal at a point)AdaptedAdapted from Ahlfors (1979), Ch. 3, §2.3, pp. 73-74; Hassani (2013), §10.3, p. 305; Riley, Hobson & Bence (2006), §24.7, pp. 839-840

Let $\Omega \subseteq \mathbb{C}$ be open, let $f : \Omega \to \mathbb{C}$ be continuous, and let $z_0 \in \Omega$ . Let $\gamma_1, \gamma_2$ be smooth arcs in $\Omega$ through $z_0$ with non-vanishing tangents there, and let $\gamma_1' , \gamma_2'$ be their images under $f$ , that is $t \mapsto f(\gamma_j(t))$ .

The map $f$ is conformal at $z_0$ if, for every such pair, the image arcs have non-vanishing tangents at $f(z_0)$ and the angle from $\gamma_1'$ to $\gamma_2'$ there equals the angle from $\gamma_1$ to $\gamma_2$ at $z_0$ , including its sign. A map is conformal on $\Omega$ if it is conformal at every point of $\Omega$ ; it is a conformal equivalence of $\Omega$ onto $\Omega_0$ if in addition it is a bijection of $\Omega$ onto $\Omega_0$ .

If the angles come out equal in size but reversed in sign, the map is called indirectly conformal; $z \mapsto \overline{z}$ is the basic example.

A clash of definitions, worth knowing about. This is the sense used by Hassani (p. 305), for whom a conformal mapping is one under which the angle at which two given curves cross is unchanged, and by Ahlfors (p. 73), who says that $w = f(z)$ is conformal at every point where $f'(z) \neq 0$ . Riley, Hobson and Bence (pp. 839-840) reserve the word for something else: for them a mapping is conformal when $w = g(z)$ and its inverse $z = h(w)$ are both analytic apart from isolated points, that is, when it is a conformal equivalence in the above language, with the isolated bad points tolerated. Their properties (i)-(iv) are then stated with the exception clause "except at points at which $g'(z)$ , and hence $h'(z)$ , is zero or infinite" attached — quoted as printed, although $h$ was introduced one page earlier as the inverse $z = h(w)$ , a function of $w$ . The two usages agree wherever $g$ is injective with $g' \neq 0$ , which is the only situation in which either is used below; this unit follows the pointwise definition and says "conformal equivalence" when bijectivity is meant.

Method(Analytic maps are conformal where the derivative does not vanish)AdaptedAdapted from Ahlfors (1979), Ch. 3, §2.3, Eqs. (1)-(2) and §2.4, Eq. (6), pp. 73-76; Riley, Hobson & Bence (2006), §24.7, Eqs. (24.26)-(24.27), p. 841

Let $f$ be analytic on an open set $\Omega$ and let $z_0 \in \Omega$ with $f'(z_0) \neq 0$ . Then:

Angles. $f$ is conformal at $z_0$ . More precisely, every smooth arc through $z_0$ with non-vanishing tangent has an image arc whose tangent direction at $f(z_0)$ is the original tangent direction turned through $\arg f'(z_0)$ — the same angle for every arc.
Lengths. $\displaystyle \lim_{z \to z_0} \frac{\lvert f(z) - f(z_0)\rvert}{\lvert z - z_0\rvert} = \lvert f'(z_0)\rvert$ , so the local change of scale is $\lvert f'(z_0)\rvert$ in every direction.
Areas. For a set $E \subset \Omega$ with area and on which $f$ is injective, $\displaystyle \operatorname{area} f(E) = \iint_E \lvert f'(z)\rvert^2\, dx\, dy$ : the local area factor is $\lvert f'(z)\rvert^2$ .
Local invertibility. $f$ is injective on some neighbourhood of $z_0$ , and the inverse is analytic there with derivative $1/f'(z_0)$ at $f(z_0)$ .

A partial converse holds. Suppose $f = u + iv$ has continuous first partial derivatives on $\Omega$ . If $f$ preserves angles at every point in the above sense, then $u$ and $v$ satisfy the Cauchy-Riemann equations, so $f$ is analytic. If instead only the isotropy of the scale change is assumed, then at each point either the Cauchy-Riemann equations hold or the conjugate equations do, so $f$ is locally either analytic or the conjugate of an analytic function. Ahlfors derives both statements from the same circle argument (p. 74).

Idea of the proofOriginal

A smooth arc is described by $t \mapsto \gamma(t)$ , and its image by $t \mapsto f(\gamma(t))$ . The chain rule gives the image tangent as a product,

\frac{d}{dt} f(\gamma(t)) = f'(\gamma(t))\, \gamma'(t) ,

and multiplication of complex numbers adds arguments and multiplies moduli. So at $z_0$ every tangent vector is turned by the one angle $\arg f'(z_0)$ and stretched by the one factor $\lvert f'(z_0)\rvert$ , no matter which arc it belonged to. A difference of two turned angles is the difference of the original two: that is conformality. The converse direction is the same identity read backwards: the most general real-linear map of the plane is $h \mapsto \alpha h + \beta \overline{h}$ , and such a map turns all directions by one angle only when $\beta = 0$ , which is the Cauchy-Riemann condition.

ProofAdaptedAdapted from Ahlfors (1979), Ch. 3, §2.3-§2.4, pp. 73-76

Write $w(t) = f(\gamma(t))$ .

Since $f$ is analytic and $\gamma$ differentiable, $w'(t_0) = f'(z_0)\gamma'(t_0)$ . Both factors are non-zero — $f'(z_0) \neq 0$ by hypothesis and $\gamma'(t_0) \neq 0$ because the arc was assumed to have a non-vanishing tangent — so $w'(t_0) \neq 0$ and the image arc has a tangent at $f(z_0)$ . Taking arguments,
$\arg w'(t_0) = \arg f'(z_0) + \arg \gamma'(t_0) \pmod{2\pi} .$
The added term does not depend on the arc. Applying this to two arcs and subtracting, the angle between the image tangents equals the angle between the original tangents, with its sign, which is claim 1.
Claim 2 is the definition of the derivative with moduli taken: $\lvert f(z) - f(z_0)\rvert / \lvert z - z_0\rvert \to \lvert f'(z_0)\rvert$ .
For claim 3, regard $f$ as the real map $(x,y) \mapsto (u,v)$ . The change-of-variables formula gives $\operatorname{area} f(E) = \iint_E \lvert u_x v_y - u_y v_x\rvert\,dx\,dy$ . By the Cauchy-Riemann equations $u_x = v_y$ and $u_y = -v_x$ , so the Jacobian determinant is $u_x^2 + u_y^2 = \lvert f'\rvert^2$ , which is already non-negative. Hence the area factor is $\lvert f'\rvert^2$ , as claimed. (This is also why a conformal map cannot reverse orientation: its Jacobian determinant is a square.)
Claim 4 is the inverse function theorem of real calculus applied to the same real map: the Jacobian determinant at $z_0$ is $\lvert f'(z_0)\rvert^2 \neq 0$ , so $f$ is a diffeomorphism of a neighbourhood of $z_0$ onto a neighbourhood of $f(z_0)$ . Writing $g$ for the inverse and differentiating $g(f(z)) = z$ along any direction shows $g$ is complex-differentiable with $g'(f(z_0)) = 1/f'(z_0)$ .

For the converse, let $h \to 0$ in $\mathbb{C}$ . Continuity of the partials makes $f$ real-differentiable, and the real derivative can always be written

f(z_0 + h) - f(z_0) = \alpha h + \beta \overline{h} + o(\lvert h\rvert), \qquad \alpha = \tfrac12\!\left(\frac{\partial f}{\partial x} - i \frac{\partial f}{\partial y}\right), \quad \beta = \tfrac12\!\left(\frac{\partial f}{\partial x} + i \frac{\partial f}{\partial y}\right) .

Take $h = \rho e^{i\theta}$ and divide by $h$ : the leading behaviour of the quotient is $\alpha + \beta e^{-2i\theta}$ , which as $\theta$ runs over $[0,2\pi)$ traces the circle of centre $\alpha$ and radius $\lvert\beta\rvert$ twice. Angle preservation requires the argument of that quotient to be independent of $\theta$ , and the argument is constant on a circle only if the circle degenerates to a point, that is $\beta = 0$ : this is $\partial f/\partial x = -i\,\partial f/\partial y$ , the Cauchy-Riemann equations in complex form. Isotropy of the scale change requires instead the modulus to be independent of $\theta$ , and the modulus is constant on a circle only if $\lvert\beta\rvert = 0$ or the centre is the origin, giving $\beta = 0$ (analytic) or $\alpha = 0$ (conjugate of analytic). $\square$

Remark(Where the hypotheses are used)AdaptedAdapted from Ahlfors (1979), Ch. 3, §2.3, pp. 74-75 and §4.1, p. 90; Riley, Hobson & Bence (2006), §24.7, p. 841 and §25.2, p. 878

$f'(z_0) \neq 0$ is what makes angles survive. It enters claim 1 twice: to give the image arc a tangent at all, and to make $\arg f'(z_0)$ a well-defined turning angle. Where $f'(z_0) = 0$ the angle is not merely undefined, it is genuinely wrong. If $f(z) - f(z_0)$ has a zero of order $m \geq 2$ at $z_0$ , then $f(z) - f(z_0) = (z-z_0)^m\,\varphi(z)$ with $\varphi(z_0) \neq 0$ , and an angle $\theta$ at $z_0$ opens into an angle $m\theta$ at $f(z_0)$ . The cleanest case is $f(z) = z^2$ at the origin: $\lvert w \rvert = \lvert z\rvert^2$ and $\arg w = 2 \arg z$ , so the first quadrant becomes the upper half-plane and the right angle at the origin is flattened to $\pi$ . Riley, Hobson and Bence make exactly this point (p. 878) about the maps $w = z^2$ and $w = z^{1/2}$ they use: because $g'$ is $0$ or $\infty$ at the origin, no angle statement is violated there. Riley, Hobson and Bence call the points where $g' = 0$ the critical points of the transformation (p. 841), because the rotation angle $\arg g'$ is undefined there.
Conformality is local; it is not injectivity. $f'$ can be non-zero throughout $\Omega$ and $f$ still fold $\Omega$ over itself, as $f(z) = z^2$ does on the annulus $1 < \lvert z\rvert < 2$ , or $f(z) = e^z$ on any horizontal strip of height more than $2\pi$ . Claim 4 gives injectivity only on some neighbourhood of each point; Ahlfors illustrates the global failure with a region whose image covers part of itself twice (p. 75), the picture that motivates a Riemann surface.
The sign of the angle needs analyticity, not merely conjugate-analyticity. $z \mapsto \overline{z}$ preserves every angle in size and reverses every one in sense, and it preserves lengths exactly. It fails the definition above, and it is indeed not analytic. This is why the partial converse has two branches: isotropy of scale alone cannot tell $f$ from $\overline{f}$ .
Continuity of the partials is needed for the converse, not for the forward direction. Without some regularity, pointwise angle preservation at a single point is too weak to force differentiability, which is why Ahlfors assumes $\partial f/\partial x$ and $\partial f /\partial y$ continuous before running the circle argument.

Definition(Mobius transformation)AdaptedAdapted from Ahlfors (1979), Ch. 3, §3.1, Eq. (7), pp. 76-78; Hassani (2013), §10.3, Eq. (10.8), p. 306

A Mobius transformation (Ahlfors: linear transformation; Hassani: homographic transformation) is a map of the extended plane $\widehat{\mathbb{C}} = \mathbb{C}\cup\{\infty\}$ of the form

S(z) = \frac{az+b}{cz+d}, \qquad a,b,c,d \in \mathbb{C}, \qquad ad - bc \neq 0 ,

with the conventions $S(\infty) = a/c$ and $S(-d/c) = \infty$ when $c \neq 0$ , and $S(\infty) = \infty$ when $c = 0$ . These are the values the formula tends to, so $S$ is continuous on $\widehat{\mathbb{C}}$ with its spherical topology.

Three facts follow at once from the formula.

Bijectivity. $S$ has the inverse $S^{-1}(w) = (dw-b)/(-cw+a)$ , again of the same form with the same non-zero determinant, so $S$ is a bijection of $\widehat{\mathbb{C}}$ onto itself. Away from $z = -d/c$ and $z = \infty$ it is analytic with $S'(z) = (ad-bc)/(cz+d)^2 \neq 0$ , hence conformal there. (The condition $ad - bc \neq 0$ is exactly what keeps $S$ from collapsing to a constant.)
Group. Composition corresponds to multiplication of the coefficient matrices $\begin{pmatrix} a & b\\ c& d\end{pmatrix}$ , so the Mobius transformations form a group; multiplying all four coefficients by the same non-zero scalar leaves $S$ unchanged, and normalising $ad-bc=1$ still leaves the choice of an overall sign.
Decomposition. If $c \neq 0$ , division gives $\frac{az+b}{cz+d} \;=\; \frac{a}{c} \;+\; \frac{bc-ad}{c^2}\cdot\frac{1}{z + d/c} ,$ so $S$ is a translation, then the inversion $z \mapsto 1/z$ , then a multiplication by a constant (a rotation composed with a scaling), then another translation. If $c = 0$ then $S(z) = (a/d)z + b/d$ and no inversion is needed.

Definition(Cross ratio)AdaptedAdapted from Ahlfors (1979), Ch. 3, §3.2, Eq. (9), p. 78

For distinct $z_2, z_3, z_4 \in \widehat{\mathbb{C}}$ , the cross ratio $(z, z_2, z_3, z_4)$ is the value at $z$ of the unique Mobius transformation carrying $z_2, z_3, z_4$ to $1, 0, \infty$ in that order. When none of the three is $\infty$ ,

(z, z_2, z_3, z_4) \;=\; \frac{z - z_3}{z - z_4} \cdot \frac{z_2 - z_4}{z_2 - z_3} ,

and when one of them is $\infty$ the two factors containing it are deleted.

Convention warning. Which of the four arguments goes to $1$ , $0$ and $\infty$ , and in which order, differs between books; the formula above is the one in Ahlfors, §3.2. Any statement below involving a cross ratio is stated for this convention, and a reader importing a formula from elsewhere should check it on the three values $z = z_2, z_3, z_4$ , where this one must give $1, 0, \infty$ .

Method(What Mobius transformations do to circles)AdaptedAdapted from Ahlfors (1979), Ch. 3, §3.2, Theorems 12-14, pp. 79-80; Hassani (2013), §10.3, Example 10.3.2, p. 306

Call a circle in the extended sense either a Euclidean circle in $\mathbb{C}$ or a Euclidean line together with the point $\infty$ . (On the Riemann sphere both are genuine circles, which is the justification for the single word.) Let $S$ be a Mobius transformation.

Invariance of the cross ratio. For any four distinct points $z_1, z_2, z_3, z_4$ of $\widehat{\mathbb{C}}$ , $(S z_1, S z_2, S z_3, S z_4) = (z_1, z_2, z_3, z_4) .$
Three points determine the map. Given distinct $z_1, z_2, z_3$ and distinct $w_1, w_2, w_3$ in $\widehat{\mathbb{C}}$ , there is exactly one Mobius transformation with $S(z_j) = w_j$ for $j = 1,2,3$ , and it is obtained by solving $(w, w_1, w_2, w_3) = (z, z_1, z_2, z_3)$ for $w$ . In particular a Mobius transformation with three fixed points is the identity.
Circles go to circles. $S$ carries every circle in the extended sense onto a circle in the extended sense. Consequently it carries each of the two complementary discs (or half-planes) bounded by such a circle onto one of the two bounded by the image circle.
Reality test. $(z_1,z_2,z_3,z_4)$ is real exactly when the four points lie on one circle in the extended sense.

Idea of the proofOriginal

Statement 2 is bookkeeping: a cross ratio is the map that normalises three points to $1, 0, \infty$ , so saying two cross ratios agree says that one normalising map followed by the inverse of the other sends each $z_j$ to $w_j$ . Statement 1 then follows because $(\,\cdot\,, Sz_2, Sz_3, Sz_4) \circ S$ normalises $z_2, z_3, z_4$ and so must be $(\,\cdot\,, z_2,z_3,z_4)$ . For statement 3 the trick is not to attack $(az+b)/(cz+d)$ at all. Instead, note that circles and lines have a single equation in $z$ and $\overline{z}$ , check that each of the three elementary pieces of the decomposition preserves the shape of that equation, and compose.

ProofAdaptedAdapted from Ahlfors (1979), Ch. 3, §3.1-§3.2, pp. 77-80; Hassani (2013), §10.3, Example 10.3.2, p. 306

Statement 2. Let $T$ be the Mobius transformation $z \mapsto (z,z_1,z_2,z_3)$ and $U$ the one $w \mapsto (w,w_1,w_2,w_3)$ ; both exist, because the decomposition in the definition shows the explicit formula is a Mobius transformation, and both are invertible. Then $S = U^{-1}T$ is a Mobius transformation with $S(z_j) = U^{-1}(T z_j) = w_j$ , because $T$ sends $z_j$ to whichever of $1, 0, \infty$ is the $j$ th, and $U$ sends $w_j$ to that same one. If $S_1, S_2$ both do this then $S_2^{-1}S_1$ fixes $z_1,z_2,z_3$ ; a Mobius transformation fixing three points has $(az+b) - z(cz+d) = 0$ , a polynomial of degree at most $2$ , with three distinct roots in $\widehat{\mathbb{C}}$ , so it vanishes identically and $S_2^{-1}S_1$ is the identity. Hence $S$ is unique.

Statement 1. Put $T z = (z, z_2, z_3, z_4)$ . The map $T S^{-1}$ is a Mobius transformation sending $Sz_2, Sz_3, Sz_4$ to $1,0,\infty$ , so by the uniqueness just proved $T S^{-1}$ is the cross-ratio map $w \mapsto (w, Sz_2, Sz_3, Sz_4)$ . Evaluating both at $w = Sz_1$ gives $(Sz_1, Sz_2,Sz_3,Sz_4) = T z_1 = (z_1,z_2,z_3,z_4)$ .

Statement 3. Every circle in the extended sense is the solution set of

A \lvert z \rvert^2 + \overline{B} z + B \overline{z} + C = 0, \qquad A, C \in \mathbb{R},\quad B \in \mathbb{C}, \quad \lvert B\rvert^2 > AC , \tag{$*$}

together with $\infty$ when $A = 0$ . Indeed, for $A \neq 0$ division by $A$ and completing the square turn $(*)$ into $\lvert z + B/A\rvert^2 = (\lvert B\rvert^2 - AC)/A^2$ , a circle of positive radius; for $A = 0$ the equation is $2\operatorname{Re}(\overline{B}z) + C = 0$ with $B \neq 0$ , a line. Conversely every circle and every line can be written this way.

Now take the three elementary maps of the decomposition in turn and substitute.

Translation $w = z + \beta$ , i.e. $z = w - \beta$ . Expanding $(*)$ regroups into $A\lvert w\rvert^2 + \overline{B'}w + B'\overline{w} + C' = 0$ with $B' = B - A\beta$ and $C' = C - 2\operatorname{Re}(\overline{B}\beta) + A\lvert\beta\rvert^2$ , both of the required type ( $C'$ real). The quantity $\lvert B'\rvert^2 - AC'$ is unchanged by a direct computation, so it stays positive.
Multiplication $w = \alpha z$ with $\alpha \neq 0$ , i.e. $z = w/\alpha$ . Substituting and multiplying by $\lvert\alpha\rvert^2$ gives $A\lvert w\rvert^2 + \overline{B\alpha}\,w + B\alpha\,\overline{w} + C\lvert\alpha\rvert^2 = 0$ , again of the form $(*)$ with $B' = B\alpha$ , $C' = C\lvert\alpha\rvert^2$ , and $\lvert B'\rvert^2 - AC' = \lvert\alpha\rvert^2(\lvert B\rvert^2 - AC) > 0$ .
Inversion $w = 1/z$ , i.e. $z = 1/w$ . Substituting into $(*)$ and multiplying through by $\lvert w\rvert^2$ gives $C\lvert w\rvert^2 + B w + \overline{B}\,\overline{w} + A = 0 ,$ which is $(*)$ with $A' = C$ , $B' = \overline{B}$ , $C' = A$ ; and $\lvert B'\rvert^2 - A'C' = \lvert B\rvert^2 - AC > 0$ . So inversion preserves the family, swapping the roles of the $\lvert z\rvert^2$ coefficient and the constant — which is exactly why it can turn a circle through the origin ( $C = 0$ ) into a line ( $A' = 0$ ) and back. Hassani runs this computation in the concrete form $\lvert z - a\rvert = r$ and finds a circle unless $r = \lvert a\rvert$ , that is unless the original circle passes through the origin, in which case the image is a line.

Since $S$ is a composition of maps of these three kinds, it preserves the family $(*)$ . A circle in the extended sense separates $\widehat{\mathbb{C}}$ into two connected pieces; $S$ is a homeomorphism of $\widehat{\mathbb{C}}$ , so it maps those two pieces homeomorphically onto the two pieces cut out by the image circle, which gives the last sentence of statement 3.

Statement 4. By statement 2 choose $S$ with $S z_2, S z_3, S z_4 = 1, 0, \infty$ ; then $(z_1,z_2,z_3,z_4) = S z_1$ by definition, and by statement 1 the question "is the cross ratio real?" is the question "does $Sz_1$ lie on $\mathbb{R}\cup\{\infty\}$ ?". The four points $z_1,z_2,z_3,z_4$ lie on a common circle in the extended sense if and only if their images do, by statement 3 applied to $S$ and to $S^{-1}$ ; and $Sz_2, Sz_3, Sz_4 = 1,0,\infty$ already determine that circle to be $\mathbb{R}\cup\{\infty\}$ , since three distinct points lie on exactly one circle in the extended sense. So the four points are concyclic exactly when $S z_1 \in \mathbb{R}\cup\{\infty\}$ , which is the claim. $\square$

Method(Harmonic functions ride along)AdaptedAdapted from Riley, Hobson & Bence (2006), §24.7, property (iv) and Eqs. (24.28)-(24.29), p. 841, and §25.2, p. 876; Hassani (2013), §10.3, Proposition 10.3.1(b), p. 306

Let $g : \Omega \to \Omega_0$ be analytic, with $\Omega, \Omega_0 \subseteq \mathbb{C}$ open, and let $\Phi_0 : \Omega_0 \to \mathbb{R}$ be twice continuously differentiable. Then

\Delta\big(\Phi_0 \circ g\big)(z) \;=\; \lvert g'(z)\rvert^2 \, \big(\Delta \Phi_0\big)(g(z)) \qquad \text{for all } z \in \Omega ,

where $\Delta = \partial_x^2 + \partial_y^2$ . In particular:

If $\Phi_0$ is harmonic on $\Omega_0$ , then $\Phi = \Phi_0 \circ g$ is harmonic on $\Omega$ — with no hypothesis on $g'$ at all.
If $g$ is in addition a conformal equivalence of $\Omega$ onto $\Omega_0$ , the correspondence $\Phi_0 \leftrightarrow \Phi_0 \circ g$ is a bijection between the harmonic functions on $\Omega_0$ and those on $\Omega$ .
Level sets correspond: $\{\Phi = c\} = g^{-1}\{\Phi_0 = c\}$ . So if $\Phi_0$ takes the constant value $c$ on a boundary arc $\Gamma_0 \subseteq \partial\Omega_0$ , and $g$ extends continuously to $\Gamma = g^{-1}(\Gamma_0) \subseteq \partial\Omega$ , then $\Phi$ takes the value $c$ on $\Gamma$ .
Where $\Phi_0 = \operatorname{Re} F$ for an analytic $F$ on $\Omega_0$ , we may take $\Phi = \operatorname{Re}(F\circ g)$ , and then the gradient magnitudes are related by $\lvert \nabla \Phi (z)\rvert = \lvert F'(g(z))\rvert\,\lvert g'(z)\rvert$ .

Riley, Hobson and Bence reach claim 1 by the route $F = f \circ h$ with $h = g^{-1}$ : an analytic function of an analytic function is analytic, and the real and imaginary parts of an analytic function are harmonic (their Eqs. (24.28)-(24.29)). They note separately (p. 876, footnote) that the solution being transported need not be given as the real part of an analytic function. Hassani states claim 1 as part (b) of Proposition 10.3.1, phrased as the preservation of harmonicity in passing from the coordinates $(x,y)$ to the image coordinates $(x',y')$ ; the symbol it uses for the function being transported is the same one its hypotheses give to the mapping, and what the book then puts to work in its examples is claim 1 above, with an arbitrary harmonic function carried along.

Idea of the proofOriginal

One identity does everything, and it is cleanest in the Wirtinger derivatives $\partial = \tfrac12(\partial_x - i\partial_y)$ and $\bar\partial = \tfrac12(\partial_x + i \partial_y)$ , in which $\Delta = 4\partial\bar\partial$ . Analyticity of $g$ says $\bar\partial g = 0$ and $\partial \overline{g} = 0$ : $g$ carries no $\overline{z}$ dependence. So in the chain rule for $\Phi_0 \circ g$ the mixed terms drop out, and what survives is the single factor $\lvert g'\rvert^2$ . The factor is non-negative, so it cannot create or destroy the vanishing of the Laplacian — which is why claim 1 needs nothing about $g'$ , while the inverse direction (claim 2) needs $g$ to be invertible.

ProofOriginal

Write $\partial = \tfrac12(\partial_x - i \partial_y)$ and $\bar\partial = \tfrac12(\partial_x + i\partial_y)$ , so that $\Delta = 4\,\partial\bar\partial$ (expand the product: the cross terms $\mp \tfrac{i}{4}\partial_x\partial_y$ cancel). For a function $\Psi$ of $z$ and $\overline{z}$ the chain rule along an analytic $g$ reads

\bar\partial\,(\Psi \circ g) \;=\; (\partial \Psi)(g)\, \bar\partial g \;+\; (\bar\partial\Psi)(g)\, \bar\partial \overline{g} \;=\; (\bar\partial \Psi)(g)\,\overline{g'} ,

because $\bar\partial g = 0$ (that is the Cauchy-Riemann condition, see Analyticity and the Cauchy-Riemann equations) and $\bar\partial\overline{g} = \overline{\partial g} = \overline{g'}$ . Applying the same rule to $\partial$ gives $\partial(\Psi\circ g) = (\partial\Psi)(g)\,g'$ .

Now take $\Psi = \Phi_0$ and compose the two steps:

\partial\bar\partial(\Phi_0\circ g) = \partial\big[(\bar\partial\Phi_0)(g)\,\overline{g'}\big] = (\partial\bar\partial\Phi_0)(g)\,g'\,\overline{g'} + (\bar\partial\Phi_0)(g)\,\partial\overline{g'} .

The second term vanishes: $g'$ is analytic, so $\overline{g'}$ is anti-analytic and $\partial \overline{g'} = \overline{\bar\partial g'} = 0$ . Multiplying by $4$ and using $g'\overline{g'} = \lvert g'\rvert^2$ ,

\Delta(\Phi_0\circ g) = \lvert g'\rvert^2\,(\Delta\Phi_0)\circ g ,

which is the identity. Claim 1 is immediate. For claim 2, apply claim 1 to $g$ and to the analytic inverse $g^{-1}$ (analytic by claim 4 of the conformality statement, since a conformal equivalence has $g' \neq 0$ throughout: a zero of $g'$ of order $m-1$ would make $g$ locally $m$ -to-one, contradicting injectivity); the two compositions are mutually inverse. Claim 3 is set theory: $\Phi(z) = c$ exactly when $\Phi_0(g(z)) = c$ . For claim 4, $F \circ g$ is analytic, so $\operatorname{Re}(F\circ g)$ is harmonic and equals $\Phi_0\circ g$ ; and for an analytic $F$ with $\Phi_0 = \operatorname{Re}F$ one has $\lvert\nabla\Phi_0\rvert = \lvert F'\rvert$ , because $F' = \partial_x \Phi_0 - i\,\partial_y\Phi_0$ by the Cauchy-Riemann equations. Applying this to $F \circ g$ and using $(F\circ g)' = F'(g)\,g'$ gives the stated product. $\square$

Method(Solving a two-dimensional Laplace problem by mapping)AdaptedAdapted from Riley, Hobson & Bence (2006), §25.2, pp. 876-878; Hassani (2013), §10.3, pp. 305-308

Given: a region $\Omega \subseteq \mathbb{C}$ and a function prescribed on $\partial\Omega$ ; wanted: the harmonic $\Phi$ on $\Omega$ with those boundary values. (The physical content is in Harmonic functions; here $\Omega$ is a cross-section of a cylindrically symmetric three-dimensional problem, so that the third coordinate drops out.)

Normalise the geometry. Place the origin and axes so that the boundary pieces lie along rays, lines or circles. Note where the data change value: those points will be the ones you have to track through every map.
Choose a model region whose Dirichlet problem you can already solve: the upper half-plane and the unit disk (Poisson integral formula), the horizontal strip, the sector.
Build the map as a composition of standard pieces, following the boundary rather than the interior:
- $z \mapsto z + \beta$ , $z\mapsto \alpha z$ : translate, rotate, rescale.
- $z \mapsto z^\beta$ with $\beta > 0$ ( $\arg w = \beta \arg z$ , $\lvert w\rvert = \lvert z\rvert^\beta$ ): the sector $\varphi_1 < \arg z < \varphi_2$ onto the sector $\beta\varphi_1 < \arg w < \beta\varphi_2$ . Ahlfors states the restriction as a necessary one — the mapping is one to one only if $\beta(\varphi_2-\varphi_1)\le 2\pi$ — and in that case the one sector is mapped onto the other (p. 90). Use $\beta = \pi/\alpha$ to open a wedge of opening $\alpha$ into a half-plane.
- $z \mapsto e^z$ : the strip $y_1 < \operatorname{Im}z < y_2$ with $y_2 - y_1 \leq 2\pi$ onto the sector $y_1 < \arg w < y_2$ ; and $z \mapsto \log z$ the other way.
- Mobius maps: half-plane $\leftrightarrow$ disk, disk $\leftrightarrow$ disk, and any three boundary points to any three. For the upper half-plane onto the unit disk with a chosen interior point $z_0$ going to the centre, use $w = e^{i\varphi}(z-z_0)/(z - \overline{z_0})$ , which has $\lvert w\rvert < 1$ exactly because an interior point is nearer to $z_0$ than to its reflection $\overline{z_0}$ (Riley, Hobson and Bence, p. 842); fix $\varphi$ by prescribing the image of one boundary point. Check each piece on the boundary points identified in step 1, and keep a branch of $\arg$ or $\log$ fixed once and for all.
Transport the data. The value prescribed at $\zeta \in \partial\Omega_0$ is the value prescribed at $g^{-1}(\zeta) \in \partial\Omega$ . Constants stay constant (claim 3 of the transport statement); a non-constant profile gets reparametrised, which is where a map that is simple on the boundary pays off.
Solve in the model region — by inspection if the data are piecewise constant on rays or arcs, otherwise by the Poisson integral or a known complex potential.
Pull back: $\Phi(z) = \Phi_0(g(z))$ . If you want the field as well, carry the analytic potential $F$ with $\operatorname{Re}F = \Phi_0$ across instead, and read the field magnitude off $\lvert (F\circ g)'\rvert = \lvert F'(g)\rvert\,\lvert g'\rvert$ .
Check. Verify the boundary values on each piece directly in the $z$ variable; verify harmonicity by exhibiting $\Phi$ as the real or imaginary part of something analytic; and locate the points where $g' = 0$ or $g' = \infty$ , since the field, though not the potential, typically misbehaves there.

Remark(Which regions can be reached)AdaptedAdapted from Ahlfors (1979), Ch. 6, §1.1, Theorem 1, p. 230 and Ch. 3, §3.2, p. 80; Riley, Hobson & Bence (2006), §24.7, Eq. (24.30), p. 843

The method of the previous block only needs a map to exist. Two results say when it does.

Riemann mapping theorem (Ahlfors, Ch. 6, Theorem 1, p. 230). Let $\Omega$ be a simply connected region that is not the whole plane, and let $z_0 \in \Omega$ . There is a unique analytic $f$ on $\Omega$ with $f(z_0) = 0$ and $f'(z_0) > 0$ that maps $\Omega$ one-to-one onto the disk $\lvert w\rvert < 1$ . Hence any two such regions are conformally equivalent, through the disk. The theorem is an existence statement — Ahlfors's proof takes $f$ to be the member of a normal family with the largest derivative at $z_0$ — and it hands you no formula, which is why one still composes elementary maps by hand.

Two caveats the hypotheses carry. Simply connected is essential: an annulus is not conformally equivalent to a disk, and two annuli are equivalent only when their radius ratios agree, so the conformal type of a multiply connected region has moduli. Not the whole plane is essential too: by Liouville's theorem (Liouville's theorem and the maximum modulus principle) a bounded analytic function on $\mathbb{C}$ is constant, so $\mathbb{C}$ admits no bounded injective analytic map at all.

Schwarz-Christoffel formula (Riley, Hobson and Bence, Eq. (24.30), p. 843). For a polygon the map can be written down. Given points $x_1,\dots,x_n$ on the real axis and interior angles $\phi_1,\dots,\phi_n$ ,

w \;=\; A\int_0^z (\xi - x_1)^{(\phi_1/\pi)-1}(\xi - x_2)^{(\phi_2/\pi)-1}\cdots(\xi-x_n)^{(\phi_n/\pi)-1}\,d\xi \;+\; B

sends $\{\operatorname{Im}z > 0\}$ bijectively to the inside of a closed polygon; the vertex $w_j$ is the image of $x_j$ , and the angle of the polygon at $w_j$ is $\phi_j$ . Each factor contributes a turn of $\pi - \phi_j$ — the exterior angle at that vertex — as $\xi$ passes $x_j$ along the real axis, since $\arg(\xi - x_j)$ drops from $\pi$ to $0$ there and the exponent multiplies that drop by $(\phi_j/\pi) - 1$ ; this is how the corners are produced. The constants $A$ and $B$ are a rotation-with-scaling and a translation of the polygon. The prices paid are that the $x_j$ corresponding to a prescribed polygon must be found numerically, and that the integral is rarely elementary.

ExampleOriginal

Steady temperature in a half-disk. Let

\Omega = \{\, z : \lvert z\rvert < 1,\ \operatorname{Im} z > 0 \,\}

be the cross-section of a long half-cylinder. The flat face is held at $T = 0$ (the diameter $-1 < x < 1$ ) and the curved face at $T = 100$ (the arc $\lvert z \rvert = 1$ , $\operatorname{Im}z>0$ ). Find the steady temperature $T$ inside, and its value at $z = i/2$ .

Step 1: a Mobius map to a quadrant. Try

\zeta = g_1(z) = \frac{1+z}{1-z} .

Follow the boundary. It sends $-1 \mapsto 0$ , $0 \mapsto 1$ and $1 \mapsto \infty$ ; three points of the circle $\mathbb{R}\cup\{\infty\}$ , so by statement 3 the diameter lands in $\mathbb{R}\cup\{\infty\}$ , and because $0 \mapsto 1 > 0$ the open diameter lands in the positive real axis. For the arc, put $z = e^{i\theta}$ with $0 < \theta < \pi$ and pull out a half-angle:

\frac{1+e^{i\theta}}{1-e^{i\theta}} = \frac{e^{-i\theta/2}+e^{i\theta/2}}{e^{-i\theta/2}-e^{i\theta/2}} = \frac{2\cos(\theta/2)}{-2i\sin(\theta/2)} = i\cot(\theta/2) ,

and $\cot(\theta/2) > 0$ for $0<\theta<\pi$ , so the arc lands in the positive imaginary axis. The two boundary pieces of $\Omega$ therefore go to the two edges of the first quadrant. The whole unit circle therefore goes to the imaginary axis together with $\infty$ (it contains $\pm i \mapsto \pm i$ as well), and $\Omega$ is one of the four connected pieces into which the unit circle and the real axis cut $\widehat{\mathbb{C}}$ . Since $g_1$ is a homeomorphism of $\widehat{\mathbb{C}}$ and $g_1(i/2) = 3/5 + 4i/5$ lies in the open first quadrant, $g_1$ maps $\Omega$ onto

Q = \{\,\zeta : \operatorname{Re}\zeta > 0,\ \operatorname{Im}\zeta>0\,\} .

Step 2: square the quadrant into a half-plane. $g_2(\zeta) = \zeta^2$ doubles arguments, so it maps $Q$ ( $0 < \arg\zeta<\pi/2$ ) onto the upper half-plane $0 < \arg W < \pi$ , carrying the positive real axis to the positive real axis and the positive imaginary axis to the negative real axis. Note $g_2'(0) = 0$ : the corner of $Q$ is flattened. That is harmless, because $0 \notin Q$ .

Step 3: transport the data and solve. The composite $g = g_2\circ g_1$ sends the diameter ( $T=0$ ) to $\{W > 0\}$ and the arc ( $T = 100$ ) to $\{W<0\}$ . In the upper half-plane the harmonic function with those two constant values is

T_0(W) = \frac{100}{\pi}\,\arg W \qquad (0 \le \arg W \le \pi) ,

harmonic because $\arg W = \operatorname{Im}\operatorname{Log}W$ with $\operatorname{Log}$ analytic off the non-positive real axis, and equal to $0$ on $\{W>0\}$ , to $100$ on $\{W<0\}$ by inspection.

Step 4: pull back. For $\zeta \in Q$ we have $\arg\zeta \in (0,\pi/2)$ , so $\arg(\zeta^2) = 2\arg\zeta$ with no branch ambiguity, and

\boxed{\;T(z) \;=\; \frac{200}{\pi}\,\arg\frac{1+z}{1-z}\;}

with the principal argument. In real coordinates, $\dfrac{1+z}{1-z} = \dfrac{(1+z)(1-\overline{z})}{\lvert 1-z\rvert^2} = \dfrac{1-\lvert z\rvert^2 + 2iy}{\lvert 1-z\rvert^2}$ , whose real part is positive on the disk, so

T(x,y) = \frac{200}{\pi}\arctan\!\left(\frac{2y}{1-x^2-y^2}\right), \qquad x^2+y^2<1,\ y>0 .

Step 5: the number. At $z = i/2$ : $2y = 1$ and $1 - \lvert z\rvert^2 = 3/4$ , so the argument of the arctangent is $4/3$ and

T(i/2) = \frac{200}{\pi}\arctan\frac43 = 59.0334\ldots \ ^\circ .

Through the maps: $\zeta = g_1(i/2) = (1+i/2)/(1-i/2) = (3/4+i)/(5/4) = \tfrac35 + \tfrac45 i$ , then $W = \zeta^2 = -\tfrac{7}{25} + \tfrac{24}{25}i$ , and $\arg W = \pi - \arctan(24/7) = 2\arctan(4/3) = 1.854590\ldots$ , giving $T = (100/\pi)(1.854590\ldots) = 59.0334\ldots$ , the same.

Step 6: checks.

Boundary. On $y = 0$ , $\lvert x\rvert<1$ : $\arctan 0 = 0$ , so $T = 0$ . Approaching $\lvert z\rvert = 1$ from inside with $y>0$ : the denominator $1-x^2-y^2 \to 0^+$ while $2y$ stays positive, so the arctangent tends to $\pi/2$ and $T \to 100$ . Both faces are right.
Harmonicity. $T = (200/\pi)\operatorname{Im}\operatorname{Log}\big((1+z)/(1-z)\big)$ , and $(1+z)/(1-z)$ is analytic on $\Omega$ with values in the right half-plane, where $\operatorname{Log}$ is analytic; so $T$ is the imaginary part of an analytic function, hence harmonic. A finite-difference Laplacian at $z = 0.3+0.4i$ with step $10^{-6}$ gives $1.5\times10^{-10}$ , consistent with zero at that step size.
A spot value. $T(0.3+0.4i) = (200/\pi)\arctan\!\big(0.8/(1-0.09-0.16)\big) = (200/\pi)\arctan(1.0\overline{6}) = 52.0529\ldots$ , between $0$ and $100$ as a maximum principle requires.
The corner points. $g'$ blows up at $z = 1$ and $g_2' = 0$ at $\zeta = 0$ (that is $z = -1$ ). Both are the two points where the boundary data jump, and $T$ has no limit there — as it should not, since the two faces meet at a corner carrying two different temperatures.

(The value $59.0334\ldots$ , the spot value and the finite-difference Laplacian were computed in double and in $25$ -digit arithmetic and agree.)

ApplicationOriginal

Corners decide how big the field gets. The power map is not just a device for simplifying a region; the exponent it carries is a physical prediction. Let the region near a corner be the wedge $W_\alpha = \{\,0 < \arg z < \alpha\,\}$ with both faces grounded, $\Phi = 0$ . Apply $w = z^{\pi/\alpha}$ , which opens $W_\alpha$ into the upper half-plane; the harmonic functions there vanishing on the whole real axis and growing slowest are multiples of $\operatorname{Im} w$ . Pulling back,

\Phi(z) = c\,\operatorname{Im}\big(z^{\pi/\alpha}\big) = c\,r^{\pi/\alpha}\sin\!\Big(\frac{\pi\theta}{\alpha}\Big), \qquad \lvert\nabla\Phi\rvert = \frac{c\,\pi}{\alpha}\,r^{(\pi/\alpha)-1} .

The exponent $(\pi/\alpha) - 1$ changes sign at $\alpha = \pi$ . At a convex corner ( $\alpha<\pi$ ) the field goes to zero; at a re-entrant corner ( $\alpha>\pi$ ) it blows up, and at the extreme $\alpha = 2\pi$ — the edge of a thin conducting sheet, or the tip of a crack — it grows like $r^{-1/2}$ . This is the reason high-voltage hardware has rounded edges, the reason lightning rods are pointed, and, with the displacement field replaced by stress, the inverse-square-root singularity at a crack tip that linear elastic fracture mechanics is built on. One conformal map gives the exponent in all these cases, and gives it exactly.

Where else it is used.

Electrostatics and magnetostatics in two dimensions. Riley, Hobson and Bence transport the field of a charged plane to a charged half-plane and to a right-angled wedge with $w = z^2$ and $w = z^{1/2}$ (pp. 877-878), and combine a wedge map with the method of images to place a line charge in a conducting wedge (p. 878). Hassani maps the outside of two parallel cylinders at different potentials onto an annulus with a single inversion $z' = 1/(z-c)$ , where the potential is a logarithm of the radius, and so obtains in closed form the potential around two charged cylinders held at different potentials (pp. 307-308).
Ideal fluid flow. The complex potential of a uniform stream, $F(w) = U w$ , pulled back by the Joukowski map $w = z + a^2/z$ , is the classical flow past a circular cylinder; pulled back by a Joukowski map off-centred, it is the flow past an aerofoil, which is how two-dimensional lift was first computed.
Steady heat conduction and diffusion. The temperature problem of the worked example is the same mathematics with $\Phi$ a temperature and $\nabla\Phi$ a heat flux; conformal mapping gives shape factors for conduction through awkward cross-sections.
Mathematics downstream. Conformal invariance of the Dirichlet integral $\iint\lvert\nabla\Phi\rvert^2$ — immediate from the area factor $\lvert g'\rvert^2$ cancelling the gradient factor $\lvert g'\rvert^{-2}$ — is what makes harmonic measure, hyperbolic geometry on the disk and two-dimensional critical phenomena conformally invariant theories.

Application

Dispersion relations

Adapted

Add to PDF

A response that cannot precede its cause has a susceptibility analytic in the upper half-plane; Cauchy's formula on a contour that skirts the real axis then expresses its real part as a principal-value integral of its imaginary part and conversely — the Kramers-Kronig relations, which tie refraction to absorption without any model of the medium.

Prerequisites: Principal values and indented contours, Cauchy's integral formula, Analytic continuation

NoteOriginal

What it means. Suppose a system is pushed and responds, linearly, with a memory kernel: the response now is a weighted integral of the forcing in the past. "In the past" is the whole of the physics. It says the kernel vanishes for negative arguments, and that single fact makes the Fourier transform of the kernel — the susceptibility $\tilde\chi(\omega)$ — analytic in the upper half of the complex $\omega$ -plane, because there the oscillating factor $e^{i\omega t}$ decays precisely on the half-line where the kernel lives.

Analyticity in a half-plane is a severe constraint. Cauchy's formula recovers an analytic function inside a contour from its values on the contour; take the contour to be the real axis closed by a large semicircle, and $\tilde\chi$ anywhere in the upper half-plane — in particular just above the real axis — is determined by $\tilde\chi$ on the real axis alone. Splitting that statement into real and imaginary parts gives a pair of integral relations: the dispersive part of the response at one frequency is fixed by the absorptive part at all frequencies, and conversely. In optics this is the Kramers-Kronig pair, and it says that a medium which absorbs at any frequency must refract at every frequency. No model of the medium is used; the input is causality, linearity, and enough decay.

Definition(Causal response, susceptibility, and the convention)Original

Let a scalar forcing $F$ produce a response $R$ through a linear, time-translation-invariant law with kernel $\chi$ ,

R(t) \;=\; \int_{-\infty}^{\infty} \chi(t - t')\,F(t')\, dt' .

The law is causal if $R(t)$ depends only on $\{F(t') : t' \le t\}$ , which for this form is exactly

\chi(\tau) = 0 \qquad\text{for } \tau < 0 .

The susceptibility is the transform of the kernel,

\tilde\chi(\omega) \;=\; \int_{-\infty}^{\infty}\chi(\tau)\,e^{i\omega\tau}\, d\tau \;=\; \int_{0}^{\infty}\chi(\tau)\,e^{i\omega\tau}\, d\tau ,

the second equality being causality. With this convention the inverse is $\chi(\tau) = \tfrac{1}{2\pi}\int \tilde\chi(\omega)e^{-i\omega\tau}d\omega$ , and a monochromatic forcing $F(t) = \operatorname{Re}(F_0 e^{-i\omega t})$ produces $R(t) = \operatorname{Re}(\tilde\chi(\omega)F_0 e^{-i\omega t})$ ; see The Fourier transform for the transform itself.

Convention. The time factor is $e^{-i\omega t}$ and the transform kernel $e^{+i\omega\tau}$ . This is the convention of Arfken, Weber and Harris, who write a wave as $\exp[i(kx-\omega t)]$ (p. 594), and it is the usual one in optics and in many-body physics. With it, causality puts the analyticity in the upper half-plane. The opposite convention $e^{-i\omega\tau}$ in the transform moves everything to the lower half-plane and reverses the sign of every imaginary part below; it is used in parts of the engineering literature, where the analogous object is the Laplace transform $\int_0^\infty \chi(\tau)e^{-s\tau}d\tau$ , analytic in $\operatorname{Re}s > 0$ (The Laplace transform). Check the convention before importing a sign.

Throughout, $u$ and $v$ denote the real and imaginary parts of the boundary values on the real axis: $f(x) = u(x) + iv(x)$ for real $x$ . For a susceptibility, $u$ is the dispersive (in-phase, reactive) part and $v$ the absorptive (out-of-phase, dissipative) part, since the work done per cycle by the forcing is proportional to $\omega\, v(\omega)\lvert F_0\rvert^2$ .

Application(Causality gives analyticity in the upper half-plane)Original

Let $\chi : \mathbb{R}\to\mathbb{C}$ be measurable with $\chi(\tau) = 0$ for $\tau<0$ and $\int_0^\infty \lvert\chi(\tau)\rvert\,d\tau = M < \infty$ . Write $\overline{H} = \{\omega \in \mathbb{C} : \operatorname{Im}\omega \ge 0\}$ and $H$ for its interior. Then

\tilde\chi(\omega) = \int_0^\infty \chi(\tau)e^{i\omega\tau}\,d\tau

is defined by an absolutely convergent integral for every $\omega \in \overline{H}$ , and

$\lvert \tilde\chi(\omega)\rvert \le M$ on $\overline{H}$ ;
$\tilde\chi$ is analytic on $H$ and continuous on $\overline{H}$ ;
$\displaystyle \sup\{\, \lvert\tilde\chi(\omega)\rvert : \omega \in \overline{H},\ \lvert\omega\rvert = R \,\} \longrightarrow 0$ as $R\to\infty$ ;
if moreover $\int_0^\infty \tau\lvert\chi(\tau)\rvert\,d\tau<\infty$ , then $\tilde\chi$ is continuously differentiable on $\overline{H}$ , hence Lipschitz on compact subsets of the real axis.

Claims 2, 3 and 4 are exactly the hypotheses the dispersion relation below needs. Conversely, analyticity in $H$ alone does not force causality — $\tilde\chi \equiv 1$ is analytic everywhere and corresponds to $\chi = \delta$ , which is causal only by the convention that places the delta at the origin, while $\tilde\chi(\omega) = e^{i\omega a}$ with $a<0$ is also analytic everywhere and is the manifestly acausal advance by $\lvert a\rvert$ . What the dispersion relation uses is analyticity plus the decay in claim 3, and $e^{i\omega a}$ with $a<0$ fails that decay.

Idea of the proofOriginal

On the half-line where $\chi$ lives, $\lvert e^{i\omega\tau}\rvert = e^{-\tau\operatorname{Im}\omega} \le 1$ for $\operatorname{Im}\omega\ge0$ : the oscillation becomes a decay exactly when the frequency is pushed upwards. That gives the bound and the convergence at once. Analyticity then needs only that the integrand be analytic in $\omega$ for each $\tau$ and that the integral converge well enough to interchange it with a contour integral, which is Morera's theorem plus Fubini. The decay in claim 3 is a two-piece estimate: the far tail of $\chi$ is small because $\chi$ is integrable, and the near part is small for large $\lvert\omega\rvert$ because a step function integrates against $e^{i\omega\tau}$ to something of size $1/\lvert\omega\rvert$ .

ProofOriginal

Convergence and claim 1. For $\omega \in \overline{H}$ and $\tau\ge0$ , $\lvert \chi(\tau)e^{i\omega\tau}\rvert = \lvert\chi(\tau)\rvert e^{-\tau\operatorname{Im}\omega} \le \lvert\chi(\tau)\rvert$ , which is integrable by hypothesis. So the integral converges absolutely and is bounded by $M$ .

Claim 2, continuity. If $\omega_n \to \omega$ in $\overline{H}$ then $\chi(\tau)e^{i\omega_n\tau}\to\chi(\tau)e^{i\omega\tau}$ pointwise, dominated by $\lvert\chi\rvert$ ; dominated convergence gives $\tilde\chi(\omega_n)\to\tilde\chi(\omega)$ .

Claim 2, analyticity. Let $\Delta$ be any closed triangle contained in $H$ , and let $\partial\Delta$ be its boundary. The function $(\omega,\tau)\mapsto \chi(\tau)e^{i\omega\tau}$ is absolutely integrable on $\partial\Delta\times(0,\infty)$ , since $\lvert\chi(\tau)e^{i\omega\tau}\rvert\le\lvert\chi(\tau)\rvert$ and $\partial\Delta$ has finite length. Fubini's theorem therefore allows the order of integration to be exchanged:

\oint_{\partial\Delta}\tilde\chi(\omega)\,d\omega = \int_0^\infty \chi(\tau)\left(\oint_{\partial\Delta} e^{i\omega\tau}\,d\omega\right) d\tau = 0 ,

because for each fixed $\tau$ the integrand $\omega\mapsto e^{i\omega\tau}$ is entire, so its integral round a closed contour vanishes by Cauchy's theorem. Since $\tilde\chi$ is continuous on $H$ and integrates to zero round every triangle there, Morera's theorem makes it analytic on $H$ .

Claim 3. Fix $\varepsilon>0$ . Choose $T$ with $\int_T^\infty\lvert\chi\rvert < \varepsilon/3$ ; then the tail of the integral is below $\varepsilon/3$ for every $\omega\in\overline{H}$ , by the same bound as in claim 1. On $[0,T]$ , choose a step function $g = \sum_{k=1}^{N}c_k\mathbf{1}_{[a_k,b_k)}$ with $\int_0^T\lvert \chi - g\rvert<\varepsilon/3$ (step functions are dense in $L^1$ of a bounded interval); the difference contributes at most $\varepsilon/3$ , again uniformly on $\overline{H}$ . Finally, for $\omega\neq0$ in $\overline{H}$ ,

\left\lvert\int_0^T g(\tau)e^{i\omega\tau}d\tau\right\rvert = \left\lvert \sum_{k=1}^{N} c_k\,\frac{e^{i\omega b_k}-e^{i\omega a_k}}{i\omega}\right\rvert \;\le\; \frac{2}{\lvert\omega\rvert}\sum_{k=1}^{N}\lvert c_k\rvert ,

using $\lvert e^{i\omega a}\rvert \le 1$ on $\overline{H}$ . The right-hand side is below $\varepsilon/3$ once $\lvert\omega\rvert$ is large enough, with a threshold depending on $g$ but not on $\arg\omega$ . Adding the three estimates gives $\lvert\tilde\chi(\omega)\rvert<\varepsilon$ for all $\omega\in\overline{H}$ with $\lvert\omega\rvert$ large, which is the uniform statement.

Claim 4. Under $\int_0^\infty\tau\lvert\chi\rvert<\infty$ the differentiated integral $\int_0^\infty i\tau\chi(\tau)e^{i\omega\tau}d\tau$ converges absolutely and uniformly on $\overline{H}$ , so it may be differentiated under the integral sign and the derivative is continuous on $\overline{H}$ by the argument of claim 2. A continuously differentiable function is Lipschitz on compacta. $\square$

Application(The dispersion relation)AdaptedAdapted from Arfken, Weber & Harris (2013), §12.8, Eqs. (12.116)-(12.118) and Fig. 12.6, pp. 591-593

Let $f$ be analytic in the open upper half-plane and continuous on the closed upper half-plane, let $f$ be Lipschitz on a neighbourhood of the real point $x_0$ , and suppose

\sup\{\lvert f(z)\rvert : \operatorname{Im}z\ge0,\ \lvert z\rvert = R\} \to 0 \qquad (R\to\infty) .

(Arfken, Weber and Harris take $f$ to be analytic above the axis and on it, and ask of the behaviour at large radius only that it be rapid enough to make the semicircular part of their Fig. 12.6 negligible. The uniform decay displayed above is one convenient condition that does this; the proof below shows it is enough.) Then, writing $f(x) = u(x)+iv(x)$ for real $x$ and $\mathrm{P}\!\int$ for the Cauchy principal value (Principal values and indented contours):

Cauchy's formula on the axis. $f(x_0) \;=\; \frac{1}{i\pi}\,\mathrm{P}\!\int_{-\infty}^{\infty}\frac{f(x)}{x-x_0}\,dx .$
The Hilbert-transform pair. Separating real and imaginary parts, $u(x_0) \;=\; \frac{1}{\pi}\,\mathrm{P}\!\int_{-\infty}^{\infty}\frac{v(x)}{x-x_0}\,dx , \qquad v(x_0) \;=\; -\frac{1}{\pi}\,\mathrm{P}\!\int_{-\infty}^{\infty}\frac{u(x)}{x-x_0}\,dx .$ These are the dispersion relations: each part of $f$ on the real axis is an integral transform of the other. The transform is the Hilbert transform, and apart from a sign it is its own inverse — which is the content of the two formulas taken together.
Off the axis. For $\operatorname{Im}z_0>0$ the same hypotheses give $f(z_0) = \frac{1}{2\pi i}\int_{-\infty}^{\infty} f(x)\,dx/(x-z_0)$ , with an ordinary integral and the full $2\pi i$ ; for $\operatorname{Im}z_0<0$ the right-hand side is $0$ .

Two remarks on reading the formulas. They say nothing unless $f$ is genuinely complex on the real axis: if $v\equiv0$ then statement 2 forces $u\equiv0$ as well, so a function that is real on the real axis and satisfies these hypotheses vanishes identically. And $u$ at the single point $x_0$ is fixed by $v$ at every frequency — the relations are global, not local.

Idea of the proofOriginal

Apply Cauchy's integral formula to $f$ on the closed semicircular contour: the real segment $[-R,R]$ together with the arc of radius $R$ in the upper half-plane. Two things then happen as $R\to\infty$ . The arc dies, because the integrand is $f(z)/(z-z_0)$ and the arc contributes at most (length) $\times$ (bound) $\approx \pi R \cdot \sup\lvert f\rvert / R$ , so uniform decay of $f$ is enough with no extra rate. And if the evaluation point is pushed down onto the real axis, the pole of the Cauchy kernel lands on the contour; indenting around it with a small semicircle costs exactly half a residue, which is why $2\pi i$ in statement 3 becomes $i\pi$ in statement 1. Everything else is taking real and imaginary parts of one equation.

ProofAdaptedAdapted from Arfken, Weber & Harris (2013), §12.8, pp. 591-592; Hassani (2013), §12.3.2, Eq. (12.6), p. 377

Fix $R > \lvert x_0\rvert + 1$ and $\delta \in (0,1)$ . Let $\Gamma_{R,\delta}$ be the closed contour, traversed counterclockwise, consisting of

the real segments $[-R,\,x_0-\delta]$ and $[x_0+\delta,\,R]$ ;
the small semicircle $C_\delta$ of radius $\delta$ about $x_0$ , run from $x_0-\delta$ to $x_0+\delta$ through the upper half-plane — so the contour steps over $x_0$ rather than through it, and this small arc is itself traversed clockwise even though the whole contour is counterclockwise;
the large arc $A_R$ : $z = Re^{i\theta}$ , $0\le\theta\le\pi$ .

Step 1: the integrand is analytic inside. The function $z\mapsto f(z)/(z-x_0)$ is analytic on the open region enclosed by $\Gamma_{R,\delta}$ , because $f$ is analytic there and the only possible singularity, $z = x_0$ , has been excluded by the indentation. By Cauchy's theorem,

\oint_{\Gamma_{R,\delta}}\frac{f(z)}{z-x_0}\,dz = 0 .

Step 2: the large arc contributes nothing. On $A_R$ we have $\lvert z - x_0\rvert \ge R - \lvert x_0\rvert$ , so

\left\lvert \int_{A_R}\frac{f(z)}{z-x_0}\,dz\right\rvert \;\le\; \pi R \cdot \frac{\sup_{A_R}\lvert f\rvert}{R-\lvert x_0\rvert} \;=\; \frac{\pi R}{R - \lvert x_0\rvert}\,\sup_{A_R}\lvert f\rvert \;\longrightarrow\; 0 ,

since the prefactor tends to $\pi$ and the supremum to $0$ by hypothesis. This is the step the decay hypothesis is for, and it shows that mere uniform decay to zero suffices: no rate is needed, because the kernel already supplies one power of $R$ .

Step 3: the indentation contributes half a residue. Lipschitz continuity of $f$ at $x_0$ lets us write $f(z) = f(x_0) + O(\lvert z-x_0\rvert)$ there, so

\int_{C_\delta}\frac{f(z)}{z-x_0}\,dz = f(x_0)\int_{C_\delta}\frac{dz}{z-x_0} + O(\delta) .

Parametrising $z = x_0+\delta e^{i\theta}$ with $\theta$ running from $\pi$ down to $0$ (so that the arc goes above $x_0$ , from left to right) gives $\int_{C_\delta} dz/(z-x_0) = \int_\pi^0 i\,d\theta = -i\pi$ . Hence the indentation tends to $-i\pi f(x_0)$ as $\delta\to0$ . This is the standard half-residue result established in Principal values and indented contours.

Step 4: the remaining segments give the principal value. By definition $\int_{-R}^{x_0-\delta} + \int_{x_0+\delta}^{R} \to \mathrm{P}\!\int_{-\infty}^{\infty} f(x)\,dx/(x-x_0)$ as $\delta\to0$ and $R\to\infty$ , provided that limit exists; the Lipschitz condition makes the symmetric deletion converge near $x_0$ (the odd part of $1/(x-x_0)$ cancels against the constant term of $f$ ), and Step 2's bound applied to $\lvert x\rvert>R$ controls the tails.

Adding the three pieces and letting $\delta\to0$ , $R\to\infty$ in Step 1 gives

\mathrm{P}\!\int_{-\infty}^{\infty}\frac{f(x)}{x-x_0}\,dx \;-\; i\pi f(x_0) \;=\; 0 ,

which is statement 1. For statement 2, substitute $f = u+iv$ and multiply by $1/(i\pi) = -i/\pi$ :

u(x_0)+iv(x_0) = \frac{1}{\pi}\mathrm{P}\!\int\frac{v(x)}{x-x_0}dx \;-\; \frac{i}{\pi}\mathrm{P}\!\int\frac{u(x)}{x-x_0}dx ,

and $u, v$ are real on the real axis, so the two principal-value integrals are real and may be matched part by part. Statement 3 is Cauchy's integral formula on the same contour without any indentation: for $\operatorname{Im}z_0>0$ the point is interior and the formula gives $2\pi i f(z_0)$ , while for $\operatorname{Im}z_0<0$ the integrand is analytic inside and the integral vanishes. $\square$

Hassani records the same computation with the choice left open: putting the small semicircle in the upper half-plane gives the upper signs in his Eq. (12.6), and putting it in the lower half-plane gives the lower signs, the two differing by the full residue (p. 377).

Application(Crossing symmetry and the Kramers-Kronig form)AdaptedAdapted from Arfken, Weber & Harris (2013), §12.8, Symmetry Relations, Eqs. (12.119)-(12.123), pp. 593-594

Suppose in addition that $f$ satisfies the crossing condition

f(-x) = f^{*}(x) \qquad (x\in\mathbb{R}) ,

equivalently that $u$ is even and $v$ is odd. (Arfken, Weber and Harris note that this holds exactly when $f$ is the transform of a real function, and that in quantum-mechanical scattering the relations $u(-x)=u(x)$ , $v(-x)=-v(x)$ are called the crossing conditions.) Then the dispersion relations can be folded onto the positive half-line, where alone a frequency is physically measurable:

u(x_0) \;=\; \frac{2}{\pi}\,\mathrm{P}\!\int_{0}^{\infty}\frac{x\,v(x)}{x^2-x_0^2}\,dx , \qquad v(x_0) \;=\; -\frac{2 x_0}{\pi}\,\mathrm{P}\!\int_{0}^{\infty}\frac{u(x)}{x^2-x_0^2}\,dx .

For the complex index of refraction these are the Kramers-Kronig relations. Since $n^2$ tends to $1$ rather than to $0$ at large frequency, the decay hypothesis is met not by $n^2$ but by $f = n^2-1$ , and the relations then read (Arfken, Weber and Harris, Eqs. (12.126))

\operatorname{Re}\big[n^2(\omega_0)-1\big] = \frac{2}{\pi}\,\mathrm{P}\!\int_0^\infty \frac{\omega\operatorname{Im}[n^2(\omega)-1]}{\omega^2-\omega_0^2}\,d\omega , \qquad \operatorname{Im}\big[n^2(\omega_0)-1\big] = -\frac{2}{\pi}\,\mathrm{P}\!\int_0^\infty\frac{\omega_0 \operatorname{Re}[n^2(\omega)-1]}{\omega^2-\omega_0^2}\,d\omega .

Knowing the absorption at all frequencies therefore fixes the refraction at every frequency, and conversely.

ProofAdaptedAdapted from Arfken, Weber & Harris (2013), §12.8, Eqs. (12.121)-(12.123), p. 593

Split the first dispersion relation at the origin and reverse the sign of the variable on the negative half:

\pi\,u(x_0) = \mathrm{P}\!\int_{-\infty}^{0}\frac{v(x)}{x-x_0}\,dx + \mathrm{P}\!\int_0^\infty\frac{v(x)}{x-x_0}\,dx = \mathrm{P}\!\int_0^\infty\frac{v(-x)}{-x-x_0}\,dx + \mathrm{P}\!\int_0^\infty\frac{v(x)}{x-x_0}\,dx .

By the crossing condition $v$ is odd, so $v(-x)/(-x-x_0) = -v(x)/(-(x+x_0)) = v(x)/(x+x_0)$ , and the two integrals combine:

\pi\,u(x_0) = \mathrm{P}\!\int_0^\infty v(x)\left[\frac{1}{x+x_0}+\frac{1}{x-x_0}\right]dx = \mathrm{P}\!\int_0^\infty v(x)\,\frac{2x}{x^2-x_0^2}\,dx ,

which is the first formula. For the second, the same splitting applied to $v(x_0) = -\frac1\pi\mathrm{P}\!\int u/(x-x_0)$ with $u$ even gives $u(-x)/(-x-x_0) = -u(x)/(x+x_0)$ , so the bracket is $1/(x-x_0)-1/(x+x_0) = 2x_0/(x^2-x_0^2)$ and the sign in front is retained. $\square$

Remark(Where the hypotheses are used)AdaptedAdapted from Arfken, Weber & Harris (2013), §12.8, pp. 591-594; Hassani (2013), §12.3.2, Eq. (12.7), p. 377; Reed & Simon (1980), §V.6C, p. 156

Analyticity in the whole open upper half-plane is what Step 1 of the proof needs; it is the mathematical content of causality. A pole of $\tilde\chi$ at $\omega_p$ with $\operatorname{Im}\omega_p>0$ would, through the inverse transform, contribute $e^{-i\omega_p t}$ to $\chi(t)$ , whose modulus is $e^{+t\operatorname{Im}\omega_p}$ : a response growing without bound, and present before the forcing. Instability and acausality are the same defect seen from two sides.
The decay hypothesis is used only in Step 2, and only to kill the large arc. It is the hypothesis that fails most often in practice, and there are two standard repairs. Subtract a constant: Arfken, Weber and Harris apply the relations to $n^2-1$ rather than $n^2$ because $n^2 \to 1$ (p. 594). Subtract at a point: Hassani's dispersion relation with one subtraction inserts an extra power of the variable in the denominator; for $\operatorname{Im}f$ odd it gives (his Eq. (12.7)) $\operatorname{Re}f(x_0) = \operatorname{Re}f(0) + \frac{2x_0^2}{\pi}\,\mathrm{P}\!\int_0^\infty \frac{\operatorname{Im}f(x)}{x\,(x^2-x_0^2)}\,dx ,$ which converges under weaker decay at the price of needing one measured value, $\operatorname{Re}f(0)$ . In scattering theory subtractions are routine; Reed and Simon say explicitly that the unsubtracted relation assumes the amplitude tends to zero, which "in general is false", and that one then modifies the argument by making subtractions (p. 156).
Lipschitz continuity at $x_0$ is used in Steps 3 and 4. Without some regularity the symmetric deletion need not converge and the principal value need not exist, so the left-hand side of statement 1 would be meaningless. It is supplied automatically by claim 4 of the causality statement whenever $\int_0^\infty\tau\lvert\chi(\tau)\rvert\,d\tau<\infty$ .
The principal value cannot be dropped. The kernel $1/(x-x_0)$ is not integrable at $x_0$ ; what converges is the symmetric deletion, and the $-i\pi f(x_0)$ of Step 3 is precisely the piece that the symmetric deletion leaves behind. Choosing instead to pass below $x_0$ changes the indentation to $+i\pi f(x_0)$ and adds a full residue to the closed-contour relation, exactly as in Principal values and indented contours; the principal value itself is the same either way.
The crossing condition is an extra assumption, not a consequence. It holds when $\chi$ is real-valued, which it is for a real response to a real forcing, and it is what allows negative frequencies — physically inaccessible — to be eliminated. A complex-valued kernel (a circularly polarised response, for instance) satisfies the full-line relations but not the folded ones.
A caution about the converse. Analyticity in the upper half-plane plus the decay of claim 3 is sufficient for the relations, but the relations do not by themselves identify a causal kernel: statement 2 is satisfied by $f\equiv0$ , and more generally two functions with the same $v$ differ by a function with $v\equiv0$ , which under the hypotheses must vanish. It is the hypotheses, not the integral identities, that carry the information.

Remark(The same argument across a cut)AdaptedAdapted from Hassani (2013), §12.3.2, pp. 376-377 and Theorem 12.3.6, pp. 375-376; Reed & Simon (1980), §V.6C, pp. 156-157

There is a second route to a dispersion relation which does not use a half-plane at all, and it is the form in which scattering theory meets the subject. Suppose $f$ is analytic on the whole plane except for a cut along the real axis from $x_0$ to $+\infty$ , and real on the real axis to the left of the cut. The Schwarz reflection principle (Analytic continuation; Hassani, Theorem 12.3.6) then gives $f(z^*) = f^*(z)$ , so the jump across the cut is purely imaginary:

f(x+i\epsilon) - f(x-i\epsilon) = 2i\operatorname{Im}f(x+i\epsilon) .

Cauchy's formula on a contour that runs along both lips of the cut and closes with a large circle — which again contributes nothing if $f$ decays — collapses to a single integral over the cut, and Hassani obtains

f(z) = \frac{1}{\pi}\int_{x_0}^{\infty}\frac{\operatorname{Im}f(x+i\epsilon)}{x-z}\,dx ,

valid for every $z$ off the cut: the function everywhere is determined by its imaginary part on the upper lip. Reed and Simon run exactly this argument for the forward scattering amplitude $A(s,0)$ , which for a wide class of potentials arises as the limit from above of a function analytic off the cut in the $s$ -plane; the jump $D = \operatorname{Im}A(s+i0,0)$ is called the discontinuity, the resulting formula is their dispersion relation (V.7a), and iterating the construction in the momentum-transfer variable produces the Mandelstam representation. They are careful to say that the relation is proved only for restricted classes of potentials, sums of Yukawa potentials among them (pp. 156-157).

The half-plane form and the cut form are the same theorem in different coordinates: the cut is where the function is not real, hence where $\operatorname{Im}f$ lives, and it plays the role the real axis plays above.

Method(Getting a dispersion relation for a given response)Original

Fix the convention. Write the time factor and the transform kernel explicitly, as in the definition block. Everything below depends on which half-plane carries the analyticity.
Check causality and integrability. Identify the kernel $\chi(\tau)$ and verify $\chi(\tau)=0$ for $\tau<0$ and $\int_0^\infty(1+\tau)\lvert\chi(\tau)\rvert\,d\tau<\infty$ . If the kernel is not known explicitly, verify instead that $\tilde\chi$ is analytic in the open upper half-plane — for a rational $\tilde\chi$ , this is the statement that every pole has negative imaginary part.
Check the decay. Examine $\tilde\chi(\omega)$ as $\lvert\omega\rvert\to\infty$ with $\operatorname{Im}\omega\ge0$ . If it tends to a non-zero constant $c$ , apply everything to $\tilde\chi - c$ ; if it does not tend to a limit at all, use the subtracted relation and be prepared to supply one measured value.
Check the symmetry. If the kernel is real-valued, $\tilde\chi(-\omega) = \tilde\chi^*(\omega)$ for real $\omega$ , so the real part is even, the imaginary part is odd, and the folded half-line relations apply. Otherwise stay on the full line.
Write the relation and, before trusting it, test it at one frequency: evaluate the principal-value integral numerically (split the interval symmetrically about the singular point, so that the two halves cancel) and compare with the closed-form value of the other part. A sign error in the convention shows up immediately as an answer of the wrong sign.
Extract a sum rule. Expand the relation for $x_0\to\infty$ . The leading term of $u(x_0)$ at large $x_0$ equals $-x_0^{-2}\cdot\frac{2}{\pi}\int_0^\infty x\,v(x)\,dx$ , so the large-frequency behaviour of the dispersive part, which is usually known from first principles, fixes a weighted integral of the absorptive part. These are the quantum-mechanical sum rules that Arfken, Weber and Harris mention at Eq. (12.123).

ExampleOriginal

The damped oscillator, checked numerically. Take a unit mass on a spring with linear damping, driven by a force $F$ :

\ddot x + \gamma\dot x + \omega_0^2 x = F(t), \qquad \gamma>0 .

The response is $x = \chi * F$ with $\chi$ the causal Green function of the operator (Causal Green's functions for initial-value problems), and transforming with $e^{i\omega t}$ ,

\tilde\chi(\omega) = \frac{1}{\omega_0^2-\omega^2-i\gamma\omega} .

Work with $\omega_0 = 1$ and $\gamma = 2/5$ , in units where $\omega_0=1$ .

Step 1: the kernel, and causality. Let $\omega_1 = \sqrt{\omega_0^2-\gamma^2/4} = \sqrt{0.96} = 0.9797958\ldots$ The kernel is

\chi(\tau) = \frac{e^{-\gamma\tau/2}}{\omega_1}\sin(\omega_1\tau)\quad (\tau>0), \qquad \chi(\tau)=0\quad(\tau<0) ,

and a direct integration confirms the transform:

\int_0^\infty \frac{e^{-\gamma\tau/2}\sin(\omega_1\tau)}{\omega_1}\,e^{i\omega\tau}\,d\tau = \frac{1}{(\gamma/2 - i\omega)^2+\omega_1^2} = \frac{1}{\omega_0^2-\omega^2-i\gamma\omega} ,

the last step because $\gamma^2/4 + \omega_1^2 = \omega_0^2$ . Both $\int_0^\infty\lvert\chi\rvert$ and $\int_0^\infty\tau\lvert\chi\rvert$ are finite because of the factor $e^{-\gamma\tau/2}$ , so all four claims of the causality statement hold.

Step 2: analyticity and decay. The denominator vanishes where $\omega^2+i\gamma\omega-\omega_0^2=0$ , that is at

\omega = \pm\omega_1 - \tfrac{i\gamma}{2} = \pm 0.9797958\ldots - 0.2\,i ,

both strictly in the lower half-plane, so $\tilde\chi$ is analytic on $\operatorname{Im}\omega\ge0$ . And $\tilde\chi(\omega) \sim -1/\omega^2$ , so the decay hypothesis holds comfortably.

Step 3: symmetry. For real $\omega$ , $\tilde\chi(-\omega) = (\omega_0^2-\omega^2+i\gamma\omega)^{-1} = \tilde\chi^*(\omega)$ : the crossing condition. So with

u(\omega) = \frac{\omega_0^2-\omega^2}{(\omega_0^2-\omega^2)^2+\gamma^2\omega^2}, \qquad v(\omega) = \frac{\gamma\omega}{(\omega_0^2-\omega^2)^2+\gamma^2\omega^2} ,

$u$ is even, $v$ is odd, and the folded relations apply.

Step 4: the test. Evaluate at $\omega_0' = 1/2$ . The denominator is $(1-\tfrac14)^2+\tfrac{4}{25}\cdot\tfrac14 = \tfrac{9}{16}+\tfrac{1}{25} = \tfrac{241}{400}$ , so exactly

u(\tfrac12) = \frac{3/4}{241/400} = \frac{300}{241} = 1.2448132780\ldots, \qquad v(\tfrac12) = \frac{1/5}{241/400} = \frac{80}{241} = 0.3319502075\ldots

Now compute the right-hand side of the first Kramers-Kronig formula numerically, splitting the integral symmetrically about $\omega=\tfrac12$ so that the two sides of the pole cancel:

\frac{2}{\pi}\,\mathrm{P}\!\int_0^\infty\frac{\omega\,v(\omega)}{\omega^2-\tfrac14}\,d\omega = 1.24481327806\ldots

against the exact $300/241 = 1.24481327801\ldots$ : agreement to ten digits, the residue being the quadrature error. The second formula, tested the same way, gives

-\frac{2\cdot\tfrac12}{\pi}\,\mathrm{P}\!\int_0^\infty\frac{u(\omega)}{\omega^2-\tfrac14}\,d\omega = 0.33195020728\ldots

against the exact $80/241 = 0.33195020747\ldots$ , agreeing to nine digits.

Step 5: the sum rule. At large $\omega_0'$ , $u(\omega_0')\to-1/\omega_0'^2$ while the folded relation gives $u(\omega_0')\to -\omega_0'^{-2}\cdot\frac{2}{\pi}\int_0^\infty \omega\,v(\omega)\,d\omega$ . Hence

\frac{2}{\pi}\int_0^\infty \omega\,v(\omega)\,d\omega = 1 ,

the oscillator-strength sum rule, with the $1$ on the right being $1/m$ for unit mass. Numerical integration of the left-hand side gives $1.0000000000$ , and the value is independent of $\gamma$ and of $\omega_0$ — damping redistributes absorption in frequency but cannot change its total weight. (The identity is also checked exactly by residues: $\omega v(\omega)$ is even, so the integral is half of $\int_{-\infty}^\infty \omega\operatorname{Im}\tilde\chi\,d\omega$ , and closing in the upper half-plane picks up nothing while the $1/\omega^2$ tail supplies the value.)

Step 6: what the numbers mean. At $\omega = \tfrac12$ , below resonance, $u>0$ : the mass moves in phase with the force, as a spring-dominated oscillator must. Above resonance $u$ changes sign ( $u(2) = -3/(9+\tfrac{16}{25}) = -75/241 < 0$ , mass-dominated, out of phase), and the sign change is tied by the dispersion relation to the peak of $v$ near $\omega=\omega_0$ . This is the mechanical original of anomalous dispersion: refraction turns over exactly where absorption is strongest.

(All numbers above were computed in $25$ -digit arithmetic and agree with the exact rational values $300/241$ , $80/241$ and $1$ to the digits shown.)

ApplicationOriginal

Optics: absorption determines refraction. This is where the subject began. The real and imaginary parts of the index of refraction are the phase velocity and the attenuation; the Kramers-Kronig pair says that measuring the absorption spectrum of a medium at all frequencies determines its refractive index at every frequency, with no model of the medium. Arfken, Weber and Harris derive the complex index from Maxwell's equations with a finite conductivity and show that it is precisely the conductivity — the absorption — that makes $n$ complex (pp. 594-595). Their Parseval identity $\int\lvert u\rvert^2 = \int\lvert v\rvert^2$ (p. 595) makes the trade exact: a medium with any refraction at all must absorb somewhere, and conversely. Hassani closes the loop with the optical theorem, which ties $\operatorname{Im}f$ to the total cross-section, and so recovers the original Kramers-Kronig formula for the real part of the forward amplitude in terms of $\sigma_{\text{tot}}$ (p. 378).
Measurement in practice. Ellipsometry, reflectivity and electron energy-loss spectroscopy all measure one part of a response over a finite window and reconstruct the other by a numerically evaluated principal-value integral. Because the relation is global, the unmeasured tails must be modelled, and the sum rule of the worked example is the standard consistency check on the result.
Circuits and control. The same theorem for the transfer function of a stable linear system is the Bode gain-phase relation: the phase at one frequency is an integral of the slope of the log-gain over all frequencies. It is the reason a sharp roll-off costs phase margin, and so the reason causality limits how aggressively a feedback loop can be shaped.
Scattering and many-body physics. The forward amplitude, the dielectric function, the conductivity and the single-particle Green function all satisfy relations of this type; the absorptive part is a measurable rate and the dispersive part a measurable shift, and the relation between them is often the only exact statement available. Reed and Simon's bootstrap discussion (pp. 156-157) is one instance of the programme of constraining an unknown amplitude by analyticity alone.
Magnetic resonance and acoustics. The absorption and dispersion lineshapes of nuclear magnetic resonance are a Hilbert-transform pair, which is why a spectrometer can record either and compute the other; attenuation and sound speed in a viscous or relaxing medium are related the same way.

Theorem · advanced

Pole expansions and infinite products

Adapted

Add to PDF

A function meromorphic in the whole plane can be written as a sum of its singular parts plus an entire remainder, and an entire function as a product over its zeros with exponential convergence factors; carrying both out for the sine gives the expansion of $\pi\cot\pi z$ and the product $\sin\pi z = \pi z\prod(1-z^2/n^2)$ .

Prerequisites: The residue theorem, Laurent series, Classification of isolated singularities, Taylor series of analytic functions, Liouville's theorem and the maximum modulus principle, Cauchy's integral formula, Complex numbers and elementary functions, Uniform convergence, Absolute and conditional convergence, Power series, Integrating and differentiating series term by term

NoteOriginal

What it means. A rational function can be written in two ways: as a sum of partial fractions, one per pole, and as a constant times a product of linear factors, one per zero. This unit says how much of that survives when there are infinitely many poles or zeros.

The obstruction is convergence, and in both cases the cure is the same. The naive sum $\sum_n P_n\bigl(1/(z-b_n)\bigr)$ and the naive product $\prod_n (1 - z/a_n)$ usually diverge, because their terms do not approach $0$ and $1$ fast enough. Subtracting from each term a polynomial (in the sum) or multiplying each factor by an exponential (in the product) repairs this without changing a single pole or zero. What is left over is an entire function, and it can often be pinned down — to a constant, or to nothing at all — by a symmetry argument and Liouville's theorem.

Carried out on $\sin\pi z$ , whose zeros are exactly the integers, the two constructions give the two classical formulas of the subject, and each is the logarithmic derivative or the integral of the other.

Definition(Singular part, infinite product, absolute convergence)AdaptedAdapted from Ahlfors (1979), Ch. 5, §2.1, p. 187, and §2.2, pp. 191-192

Let $f$ be meromorphic in a region $\Omega$ : analytic there except for poles. At a pole $b$ the Laurent expansion (Laurent series) has finitely many negative powers, and the singular part of $f$ at $b$ is the sum of those terms,

P\!\left(\frac{1}{z-b}\right) = \sum_{k=1}^{N}\frac{a_{-k}}{(z-b)^{k}},

a polynomial $P$ in $1/(z-b)$ with no constant term. The difference $f(z) - P(1/(z-b))$ is analytic at $b$ .

An infinite product $\prod_{n\ge1}p_n$ is evaluated through its partial products $P_N = p_1\cdots p_N$ . Following Ahlfors, we call it convergent when at most finitely many factors vanish and the partial products formed from the non-vanishing factors tend to a finite limit that is not zero. The value zero is excluded deliberately: were it allowed, a single zero factor would make any product converge, whatever the other factors did. In a convergent product $p_n \to 1$ , so products are always written

\prod_{n=1}^{\infty}(1 + a_n), \qquad a_n \to 0 .

The product is absolutely convergent when $\sum_n \log(1+a_n)$ converges absolutely, the principal branch of the logarithm being taken in every term.

An entire function is one analytic in the whole plane.

Theorem(Convergence of infinite products)AdaptedAdapted from Ahlfors (1979), Ch. 5, §2.2, Thms 5 and 6, pp. 191-192; Stein & Shakarchi (2003), Ch. 8, Prop. 1.9, p. 247

Let $1 + a_n \ne 0$ for every $n$ .

The product $\prod_{n\ge1}(1+a_n)$ converges if and only if the series $\sum_{n\ge1}\log(1+a_n)$ converges, the terms being principal values.
The product converges absolutely if and only if $\sum_{n\ge1}\lvert a_n\rvert$ converges.

Convergence of $\sum_n a_n$ is neither necessary nor sufficient for convergence of the product.

The same criteria apply to a product of functions, uniformly on a set: if $\sum_n\lvert a_n(z)\rvert$ converges uniformly on a compact set $K$ on which no factor vanishes, then $\prod_n (1+a_n(z))$ converges uniformly on $K$ .

Theorem(Mittag-Leffler: prescribing the poles)AdaptedAdapted from Ahlfors (1979), Ch. 5, §2.1, Thm 4, pp. 187-188

The data. Distinct points $b_1, b_2, \dots$ whose moduli tend to infinity, and for each $n$ a polynomial $P_n$ vanishing at the origin, $P_n(0) = 0$ .

Existence. At least one function meromorphic on all of $\mathbb{C}$ has its set of poles equal to $\{b_n\}$ and has $P_n\bigl(1/(z - b_n)\bigr)$ for its singular part at $b_n$ .

Shape of the general solution. Every function meeting that description can be exhibited as

f(z) \;=\; \sum_{n}\left[ P_n\!\left(\frac{1}{z-b_n}\right) - p_n(z)\right] \;+\; g(z),

in which the $p_n$ are polynomials picked to force convergence and $g$ is entire. Two things here are at our disposal rather than determined by the data: the $p_n$ , which may be changed for any other polynomials that still make the series converge, since the difference is entire and is absorbed into $g$ ; and $g$ itself, about which the poles say nothing whatever.

Theorem(Mittag-Leffler: the simple-pole form used in practice)AdaptedAdapted from Arfken, Weber & Harris (2013), §11.7, Eq. (11.79), pp. 515-516; Hassani (2013), §12.1, Eq. (12.1), pp. 363-364

Suppose $f$ is analytic at $z = 0$ and at every other finite point except for discrete simple poles at $z_1, z_2, \dots$ , ordered so that $0 < \lvert z_1\rvert \le \lvert z_2\rvert \le \cdots$ , with residues $b_1, b_2, \dots$ , and suppose that $\lvert f(z)/z\rvert \to 0$ in the limit of large $z$ . Then

f(z) \;=\; f(0) \;+\; \sum_{n=1}^{\infty} b_n\left[\frac{1}{z - z_n} + \frac{1}{z_n}\right].

Hassani states the same expansion with the hypothesis that $\lvert f(\xi)\rvert$ approach a finite value as $\lvert\xi\rvert\to\infty$ , which is the stronger of the two conditions and implies Arfken, Weber and Harris's. The constants $1/z_n$ are what make the series converge; they are the first Taylor coefficients of the singular parts, in the role played by the polynomials $p_n$ in the general theorem.

Theorem(The cotangent expansion)AdaptedAdapted from Ahlfors (1979), Ch. 5, §2.1, Eqs. (9), (10), (11), pp. 188-189; Arfken, Weber & Harris (2013), §11.7, Eq. (11.81), p. 518

For every $z \in \mathbb{C}$ that is not an integer,

\frac{\pi^{2}}{\sin^{2}\pi z} \;=\; \sum_{n=-\infty}^{\infty}\frac{1}{(z-n)^{2}},

\pi\cot\pi z \;=\; \frac1z + \sum_{n \ne 0}\left[\frac{1}{z-n} + \frac1n\right] \;=\; \frac1z + \sum_{n=1}^{\infty}\frac{2z}{z^{2}-n^{2}} .

Both series converge absolutely and uniformly on every compact set from which the integers have been removed. Arfken, Weber and Harris write the second one without the factor $\pi$ in the argument, as $\cot z = z^{-1} + 2z\sum_{n\ge1}(z^2 - n^2\pi^2)^{-1}$ ; the two are the same statement after $z \mapsto \pi z$ .

Theorem(Weierstrass: prescribing the zeros)AdaptedAdapted from Ahlfors (1979), Ch. 5, §2.3, Thm 7, p. 195, and the discussion of canonical products and genus, p. 196

The data. An integer $m \ge 0$ and non-zero numbers $a_1, a_2, \dots$ , repeated as often as the multiplicity wanted and with $\lvert a_n\rvert \to \infty$ ; a finite list is permitted.

Existence. Some entire function vanishes to order $m$ at the origin, vanishes at each $a_n$ to the prescribed multiplicity, and vanishes nowhere else.

Shape of the general solution. Any entire function with that zero set, and that zero set only, can be exhibited as

f(z) \;=\; z^{m}e^{g(z)}\prod_{n=1}^{\infty}\left(1 - \frac{z}{a_n}\right) \exp\!\left[\frac{z}{a_n} + \frac12\left(\frac{z}{a_n}\right)^{2} + \cdots + \frac{1}{m_n}\left(\frac{z}{a_n}\right)^{m_n}\right],

with suitable integers $m_n$ and $g$ entire.

The exponents $m_n$ may have to grow with $n$ . When they do not have to — that is, when a single integer $h$ can be used for all of them, which happens precisely for those $h$ with $\sum_n \lvert a_n\rvert^{-h-1} < \infty$ — one takes the smallest such $h$ . The product built with that constant exponent is then called the canonical product belonging to $\{a_n\}$ , and $h$ is its genus. The case $h = 0$ costs no exponential factors at all: the product collapses to $\prod_n(1 - z/a_n)$ .

Theorem(The sine product)AdaptedAdapted from Ahlfors (1979), Ch. 5, §2.3, Eqs. (23) and (24), p. 197; Arfken, Weber & Harris (2013), §11.7, Eq. (11.89), p. 520

For every $z \in \mathbb{C}$ ,

\sin \pi z \;=\; \pi z \prod_{n \ne 0}\left(1 - \frac{z}{n}\right)e^{z/n} \;=\; \pi z \prod_{n=1}^{\infty}\left(1 - \frac{z^{2}}{n^{2}}\right),

The first product is the canonical product belonging to the non-zero integers: $\sum_n 1/\lvert n\rvert$ diverges and $\sum_n 1/n^{2}$ converges, so no smaller exponent than $h = 1$ would do, and no larger one is needed. The factor standing in front is $\pi z$ , whose exponential part $e^{g}$ is the constant $\pi$ and so contributes a polynomial of degree $0$ ; the genus of $\sin\pi z$ is therefore $1$ as well.

Idea of the proofOriginal

Three ideas, used in this order.

Convergence is bought with free parameters. The singular part at $b_n$ is analytic on $\lvert z\rvert < \lvert b_n\rvert$ , so near the origin it has a Taylor series. Subtracting a long enough initial segment of that series leaves a remainder that is small on a fixed disc — as small as we like, because we may take the segment as long as we like. Choosing the lengths so that the remainders are majorised by $2^{-n}$ makes the whole sum converge, and subtracting polynomials changes no singular part. The product argument is the same statement for $\log(1 - z/a_n)$ , whose Taylor segment appears as the exponent of a convergence factor.

An identity between two such expansions is an entire function, and entire functions can be killed. Once a sum with the right poles exists, the difference between it and the function we started from is entire. To show that difference is zero, find something it cannot be: here both $\pi^{2}/\sin^{2}\pi z$ and its candidate series have period $1$ and tend to $0$ far from the real axis, so their difference is a bounded entire function with limit zero, hence zero by Liouville's theorem.

Products and pole expansions are the same statement, one derivative apart. The logarithmic derivative $f'/f$ of an entire function has a simple pole of residue $1$ at each simple zero of $f$ and nothing else, so a product formula for $f$ is a pole expansion for $f'/f$ , and conversely one integrates. That is how the sine product falls out of the cotangent expansion, and the step is reversible.

ProofAdaptedAdapted from Ahlfors (1979), Ch. 5, §2.1, pp. 188-189; Ahlfors (1979), Ch. 5, §2.2, pp. 191-192; Ahlfors (1979), Ch. 5, §2.3, pp. 194-195, 197; Arfken, Weber & Harris (2013), §11.7, pp. 515-516, 519-520

Part A: products reduce to series.

Suppose first that $\sum_n\lvert a_n\rvert$ converges. Then $a_n\to0$ , so after discarding finitely many terms we may assume $\lvert a_n\rvert \le 1/2$ throughout. For $\lvert w\rvert \le 1/2$ the principal logarithm has the convergent expansion $\log(1+w) = w - w^{2}/2 + w^{3}/3 - \cdots$ (Power series), so

\bigl\lvert \log(1+w) - w\bigr\rvert \le \frac{\lvert w\rvert^{2}}{2}\bigl(1 + \lvert w\rvert + \lvert w\rvert^{2}+\cdots\bigr) \le \lvert w\rvert^{2},

by the triangle inequality and the geometric series, whence $\tfrac12\lvert w\rvert \le \lvert\log(1+w)\rvert \le 2\lvert w\rvert$ for $\lvert w\rvert\le 1/2$ . (This is the bound Stein and Shakarchi isolate on p. 247, for real arguments.) Consequently $\sum_n\log(1+a_n)$ converges absolutely; call its sum $S$ and its partial sums $S_N$ . Since $\exp$ is continuous and $P_N = e^{S_N}$ , the partial products converge to $e^{S} \ne 0$ . Conversely, if $\sum_n\log(1+a_n)$ converges absolutely then its terms tend to $0$ , so $1+a_n = e^{\log(1+a_n)}\to1$ and $a_n\to0$ ; the same two-sided bound then gives $\lvert a_n\rvert \le 2\lvert\log(1+a_n)\rvert$ for large $n$ , and $\sum_n\lvert a_n\rvert$ converges. That is part 2.

For part 1 only the direction just proved is needed below: convergence of $\sum_n\log(1+a_n)$ to $S$ gives $P_N = e^{S_N}\to e^{S}\ne0$ . The converse is Ahlfors's Theorem 5 (p. 192); we use it only through part 2. If the bounds above hold uniformly in $z$ on a set $K$ — as they do when $\sum_n\lvert a_n(z)\rvert$ converges uniformly on $K$ — then $S_N(z)\to S(z)$ uniformly, and so does $e^{S_N(z)}$ , because $\exp$ is uniformly continuous on the bounded set where the $S_N$ live. This gives the uniform statement.

Part B: existence of a function with prescribed singular parts.

If one of the $b_n$ is $0$ , set that single term aside and add it back at the end; it is one more term in a sum that is already finite near the origin. So assume every $b_n \ne 0$ , which is what the Taylor expansion below needs. Fix $n$ and write $h_n(z) = P_n\bigl(1/(z-b_n)\bigr)$ , which is analytic on the disc $\lvert z\rvert < \lvert b_n\rvert$ . Let $M_n$ be its maximum on the circle $\lvert z\rvert = \lvert b_n\rvert/2$ , and let $p_n$ be the Taylor polynomial of $h_n$ at the origin of degree $\nu_n$ , to be chosen. By the Cauchy integral formula on that circle (Cauchy's integral formula), the $k$ -th term of the finite geometric sum $\sum_{k\le\nu_n} z^{k}\zeta^{-k-1}$ integrates against $h_n$ to the $k$ -th Taylor coefficient of $h_n$ times $z^{k}$ , so that sum accounts for $p_n$ exactly; and subtracting it from $1/(\zeta-z)$ leaves $(\zeta-z)^{-1}(z/\zeta)^{\nu_n+1}$ . Hence for $\lvert z\rvert \le \lvert b_n\rvert/4$ ,

\bigl\lvert h_n(z) - p_n(z)\bigr\rvert = \left\lvert \frac{1}{2\pi i}\oint_{\lvert \zeta\rvert = \lvert b_n\rvert/2}\frac{h_n(\zeta)}{\zeta - z}\left(\frac{z}{\zeta}\right)^{\nu_n+1}\!d\zeta\right\rvert \;\le\; 2 M_n\left(\frac{2\lvert z\rvert}{\lvert b_n\rvert}\right)^{\nu_n+1} ,

since the circle has length $2\pi\lvert b_n\rvert/2$ and $\lvert \zeta - z\rvert \ge \lvert b_n\rvert/4$ on it. Now choose $\nu_n$ so large that $2^{\nu_n} \ge M_n 2^{\,n}$ . Because $2\lvert z\rvert/\lvert b_n\rvert \le 1/2$ in the range considered, the bound becomes

\bigl\lvert h_n(z) - p_n(z)\bigr\rvert \;\le\; 2M_n\,2^{-\nu_n-1} \;=\; M_n 2^{-\nu_n} \;\le\; 2^{-n} \qquad\bigl(\lvert z\rvert \le \lvert b_n\rvert/4\bigr).

Fix $R>0$ . Since $b_n\to\infty$ , all but finitely many $n$ satisfy $\lvert b_n\rvert \ge 4R$ , and for those the displayed bound holds on the whole disc $\lvert z\rvert\le R$ . By the Weierstrass $M$ -test (Uniform convergence) the tail $\sum_{\lvert b_n\rvert\ge 4R}(h_n - p_n)$ converges absolutely and uniformly there; being a uniform limit of analytic functions it is analytic, by Morera's theorem (Cauchy's integral formula) together with termwise integration (Integrating and differentiating series term by term). Adding back the finitely many omitted terms gives a function meromorphic on $\lvert z\rvert < R$ whose poles there are the $b_n$ in that disc, with the prescribed singular parts: each omitted term contributes its own singular part, and the tail is analytic. As $R$ was arbitrary and the construction does not depend on $R$ , the series defines one meromorphic function on $\mathbb{C}$ .

Finally, if $f_1$ and $f_2$ are both meromorphic in the plane with these poles and these singular parts, then $f_1 - f_2$ has a removable singularity at every $b_n$ and is analytic elsewhere, hence is entire. This is the form displayed in the statement.

Part C: the two trigonometric pole expansions.

The double-pole series. Put $F(z) = \pi^{2}/\sin^{2}\pi z$ and $G(z) = \sum_{n\in\mathbb{Z}}(z-n)^{-2}$ . The zeros of $\sin\pi z$ are exactly the integers and are simple (Complex numbers and elementary functions), so $F$ has a double pole at each integer and no other singularity; near $n$ , $\sin^{2}\pi(z-n) = \sin^2 \pi z$ gives the singular part $(z-n)^{-2}$ , the same as the $n$ -th term of $G$ . The series $G$ converges absolutely and uniformly on any compact set once the finitely many terms that blow up on it are removed, by comparison with $\sum n^{-2}$ , so $G$ is meromorphic with exactly the same poles and singular parts. By Part B, $g = F - G$ is entire.

Both $F$ and $G$ are unchanged by $z\mapsto z+1$ , so $g$ has period $1$ and it is enough to bound $g$ on the strip $0 \le x \le 1$ , $z = x+iy$ . For $F$ : $\lvert\sin\pi z\rvert^{2} = \cosh^{2}\pi y - \cos^{2}\pi x \ge \cosh^{2}\pi y - 1$ , so $\lvert F\rvert \to 0$ uniformly in $x$ as $\lvert y\rvert\to\infty$ . For $G$ , with $0\le x\le1$ and $\lvert n\rvert \ge 2$ we have $\lvert x - n\rvert \ge \lvert n\rvert - 1$ , hence

\lvert G(z)\rvert \;\le\; \sum_{\lvert n\rvert\le1}\frac{1}{y^{2}} + 2\sum_{m\ge1}\frac{1}{m^{2}+y^{2}} \;\le\; \frac{3}{y^{2}} + 2\int_{0}^{\infty}\frac{dt}{t^{2}+y^{2}} \;=\; \frac{3}{y^{2}} + \frac{\pi}{\lvert y\rvert},

using $\lvert z-n\rvert^{2} = (x-n)^{2}+y^{2}$ and comparing the decreasing terms with the integral. So $G\to0$ as well, and $g\to0$ uniformly as $\lvert y\rvert\to\infty$ in the strip. Being continuous, $g$ is bounded on the remaining compact part of the strip, hence bounded on the strip, hence — by periodicity — bounded on $\mathbb{C}$ . Liouville's theorem (Liouville's theorem and the maximum modulus principle) makes $g$ constant, and the limit $0$ makes the constant $0$ . This proves the first displayed identity.

The cotangent series. Set $H(z) = z^{-1} + \sum_{n\ne0}\bigl[(z-n)^{-1} + n^{-1}\bigr]$ . Its $n$ -th term equals $z/\bigl(n(z-n)\bigr)$ , which for fixed $z$ is $O(n^{-2})$ , so $H$ converges absolutely and uniformly on compact sets away from the integers, and may be differentiated term by term there (Integrating and differentiating series term by term); the derivative is $-G(z)$ . By the identity just proved, $\frac{d}{dz}\bigl[\pi\cot\pi z\bigr] = -\pi^{2}/\sin^{2}\pi z = -G(z)$ too. Hence $\pi\cot\pi z - H(z)$ has vanishing derivative on $\mathbb{C}\setminus\mathbb{Z}$ , which is open and connected (removing a discrete set from the plane leaves it path-connected), so the difference is one constant there. Both functions are odd — for $H$ , because the substitution $n\mapsto -n$ carries the $n$ -th term to minus the $(-n)$ -th — so the constant equals its own negative and is $0$ . Bracketing $n$ with $-n$ , $\frac{1}{z-n}+\frac1n+\frac{1}{z+n}-\frac1n = \frac{2z}{z^{2}-n^{2}}$ , which is the second form.

Part D: the Weierstrass product.

Let $a_n \ne 0$ with $a_n\to\infty$ . For integers $m_n \ge 0$ put

E_n(z) = \left(1-\frac{z}{a_n}\right)\exp\!\left[\sum_{k=1}^{m_n}\frac1k\left(\frac{z}{a_n}\right)^{k}\right].

Fix $R$ and consider only the $n$ with $\lvert a_n\rvert > 2R$ ; since $a_n\to\infty$ there are only finitely many others, and each of those contributes a single factor, analytic with one zero. On $\lvert z\rvert\le R$ the principal logarithm of $1 - z/a_n$ is given by its Taylor series, so

r_n(z) := \log\left(1-\frac{z}{a_n}\right) + \sum_{k=1}^{m_n}\frac1k\left(\frac{z}{a_n}\right)^{k} = -\sum_{k>m_n}\frac1k\left(\frac{z}{a_n}\right)^{k},

and therefore, bounding $1/k$ by $1/(m_n+1)$ and summing the geometric series,

\lvert r_n(z)\rvert \;\le\; \frac{1}{m_n+1}\left(\frac{R}{\lvert a_n\rvert}\right)^{m_n+1}\left(1-\frac{R}{\lvert a_n\rvert}\right)^{-1}.

Take $m_n = n$ . For all $n$ with $\lvert a_n\rvert > 2R$ — all but finitely many, since $a_n\to\infty$ — the right-hand side is at most $2\cdot 2^{-(n+1)}$ , so $\sum_n\lvert r_n(z)\rvert$ converges uniformly on $\lvert z\rvert\le R$ . By Part A the product $\prod_n E_n(z) = \prod_n e^{r_n(z)}$ converges uniformly on $\lvert z\rvert\le R$ to an analytic function with no zeros there, and restoring the finitely many factors with $\lvert a_n\rvert\le 2R$ gives an analytic function on $\lvert z\rvert<R$ whose zeros are exactly the $a_n$ lying there, with the right multiplicities. As $R$ is arbitrary, $z^{m}\prod_n E_n(z)$ is entire with exactly the prescribed zeros.

If $f$ is any entire function with exactly those zeros, the quotient $q(z) = f(z)\big/\bigl(z^{m}\prod_n E_n(z)\bigr)$ is entire and never zero; then $q'/q$ is entire, so it has an entire primitive $g$ (Taylor series of analytic functions), and $\bigl(qe^{-g}\bigr)' = e^{-g}(q' - qg') = 0$ , so $q = Ce^{g}$ with $C \ne 0$ ; absorbing $C$ into $g$ gives the displayed form. If one $h$ serves as all the $m_n$ , the estimate above shows the product converges precisely when $\sum_n\lvert a_n\rvert^{-h-1} < \infty$ , which is the condition defining the genus.

Part E: the sine product.

The zeros of $\sin\pi z$ are the integers, each simple, so $m = 1$ and $\{a_n\} = \{n : n \ne 0\}$ . Here $\sum 1/\lvert n\rvert$ diverges while $\sum 1/n^{2}$ converges, so the genus is $h = 1$ and Part D gives

\sin\pi z = z\,e^{g(z)}\prod_{n\ne0}\left(1-\frac{z}{n}\right)e^{z/n}

for some entire $g$ . Take logarithmic derivatives. The product converges uniformly on compact sets, so the logarithmic derivative may be taken term by term (Integrating and differentiating series term by term) on any compact set avoiding the integers:

\pi\cot\pi z \;=\; \frac1z + g'(z) + \sum_{n\ne0}\left[\frac{1}{z-n}+\frac1n\right].

Comparing with Part C gives $g' \equiv 0$ , so $g$ is a constant. Dividing by $z$ and letting $z\to0$ , the product tends to $1$ and $\sin\pi z/z \to \pi$ , so $e^{g} = \pi$ . Pairing the factors for $n$ and $-n$ , the exponentials cancel and $(1-z/n)(1+z/n) = 1 - z^{2}/n^{2}$ , which is the second form. $\square$

Remark(Where the hypotheses are used)Original

$b_n\to\infty$ in Mittag-Leffler's theorem. It is used twice in Part B: to make all but finitely many $\lvert b_n\rvert$ exceed $4R$ , so that the tail estimate applies on $\lvert z\rvert\le R$ , and to make the pole set discrete. It cannot be dropped, and not merely for technical reasons: if the $b_n$ had a finite accumulation point $c$ , then $c$ would be a non-isolated singularity, and a function meromorphic in the whole plane has only isolated singularities (Classification of isolated singularities). Prescribing simple poles at $b_n = 1/n$ is therefore impossible in the plane, though it is possible in $\mathbb{C}\setminus\{0\}$ .
The subtracted polynomials $p_n$ are not optional. Prescribe a simple pole of residue $1$ at every integer. The bare series $\sum_n (z-n)^{-1}$ diverges at every $z$ , its terms behaving like $-1/n$ ; subtracting the constants $1/n$ , which changes no singular part, makes the terms $O(n^{-2})$ and the series converge. That is precisely the step from the first to the second form of the cotangent expansion. By contrast, for the double poles of $\pi^{2}/\sin^{2}\pi z$ no subtraction is needed, because $(z-n)^{-2}$ is already $O(n^{-2})$ .
Absolute, not conditional, convergence of $\sum a_n$ controls a product. Take $a_n = (-1)^{n}/\sqrt n$ for $n \ge 2$ . Then $\sum a_n$ converges, by the alternating series test. But $\log(1+a_n) = a_n - a_n^{2}/2 + O(n^{-3/2})$ , and $\sum a_n^{2}/2 = \sum 1/(2n)$ diverges, so the partial sums of $\sum\log(1+a_n)$ tend to $-\infty$ and the partial products tend to $0$ : the product diverges in the sense fixed above. (Numerically, the partial product through $n = 2\times10^{6}$ is $1.187\times10^{-3}$ and still falling, while $\sum_{n\ge2}a_n = 0.3951$ .) This is the "neither sufficient nor necessary" remark on Ahlfors p. 193, made concrete.
Periodicity and decay are what Liouville's theorem is fed in Part C. Periodicity converts a bound on one strip into a bound on the plane; decay as $\lvert y\rvert\to\infty$ then fixes the constant at $0$ . Drop either and the argument fails: $F - G + 1$ and $F - G + \sin 2\pi z$ both have the same poles and singular parts as $F$ , and only the extra information rules them out. The entire remainder in Mittag-Leffler's theorem is genuinely free, and every identification of it is an extra argument of this kind.
Arfken, Weber and Harris's growth hypothesis is a statement about expanding circles. Their condition reads $\lvert f(z)/z\rvert\to0$ in the limit of large $z$ , and their proof uses it only on the circles $C_N$ of radius $R_N$ enclosing the first $N$ poles: there the modulus of the integrand of $\oint_{C_N} f(w)\,dw/\bigl(w(w-z)\bigr)$ behaves like $\lvert f(R_N)\rvert/R_N^{2}$ , so the integral is at most about $2\pi\lvert f(R_N)\rvert/R_N$ , which their hypothesis sends to zero. Since $f$ has poles marching off to infinity, the bound cannot hold on arbitrary large circles, and in any application one chooses the radii $R_N$ to keep a fixed distance from the poles — for $\cot$ and $\tan$ , circles through the midpoints between consecutive poles. Ahlfors's Theorem 4 needs no growth hypothesis at all, because it does not claim that the sum of bare singular parts converges.
Genus needs a smallest $h$ to exist. The canonical product is available only when $\sum_n\lvert a_n\rvert^{-h-1}$ converges for some integer $h$ . Theorem 7, with $m_n$ allowed to vary and $m_n = n$ always working, has no such restriction, which is why it is the general statement.

Method(Getting a pole expansion or a product)Original

To expand a given meromorphic $f$ in partial fractions:

List the poles and singular parts. Residues at simple poles come from the rules in The residue theorem; higher-order singular parts come from a local Laurent expansion.
Test the bare sum. If $\sum_n P_n(1/(z-b_n))$ already converges locally uniformly off the poles — which happens when the terms are $O(\lvert b_n\rvert^{-2})$ — use it. Otherwise subtract the leading Taylor terms $p_n$ of each singular part, the constant $P_n(-1/b_n)$ being enough whenever the residues are bounded and $\sum\lvert b_n\rvert^{-2} < \infty$ .
Name the entire remainder and kill it. Write $f = (\text{series}) + g$ with $g$ entire and find a property that forces $g$ to be constant: periodicity plus decay in a strip, or a global bound, then Liouville. Evaluate at one convenient point to fix the constant.
Check against a known special value, for instance by comparing the Laurent coefficients of both sides at one pole.

To obtain a product for an entire $f$ :

Locate the zeros $a_n$ with multiplicity, and the order $m$ of the zero at the origin if there is one.
Find the genus: the smallest integer $h \ge 0$ with $\sum_n\lvert a_n\rvert^{-h-1} < \infty$ . Then write the canonical product with exponential factors $\exp\bigl[\sum_{k\le h}(z/a_n)^{k}/k\bigr]$ ; for $h = 0$ there are none.
Write $f = z^{m}e^{g}\times(\text{canonical product})$ and take logarithmic derivatives. This turns the unknown $g'$ into the entire remainder of a pole expansion of $f'/f$ , which step 3 above handles.
Fix the surviving constant by a limit at one point, usually $z\to0$ after dividing by $z^{m}$ .

Example(An entire function with zeros at the squares, and two numbers)Original

Prescribe simple zeros at $a_n = n^{2}$ , $n = 1, 2, 3, \dots$ , and no others.

Step 1: the genus. $\sum_n \lvert a_n\rvert^{-1} = \sum_n n^{-2} = \pi^{2}/6$ , which is finite, so $h = 0$ already works and is the smallest possibility. The canonical product therefore carries no exponential factors:

P(z) \;=\; \prod_{n=1}^{\infty}\left(1 - \frac{z}{n^{2}}\right).

By Part A it converges absolutely and uniformly on every disc, since $\sum_n\lvert z\rvert/n^{2}$ does.

Step 2: identify it in closed form. Put $z = w^{2}$ in the sine product: $\sin\pi w = \pi w\prod_{n\ge1}(1-w^{2}/n^{2})$ , so

P(w^{2}) \;=\; \frac{\sin\pi w}{\pi w}\qquad\text{for all } w \ne 0,

and both sides extend to the value $1$ at $w=0$ . The right-hand side is an even entire function of $w$ , so it depends only on $w^{2}$ and the formula determines $P$ at every $z$ , whichever square root of $z$ is used. In the Weierstrass form of Part D this says $f(z) = P(z)$ has $m = 0$ and $g \equiv 0$ .

Step 3: a number from the product. Take $z = -1/4$ , that is $w = i/2$ . Then

\prod_{n=1}^{\infty}\left(1 + \frac{1}{4n^{2}}\right) \;=\; P(-\tfrac14) \;=\; \frac{\sin(i\pi/2)}{i\pi/2} \;=\; \frac{\sinh(\pi/2)}{\pi/2} \;=\; 1.4650523833 .

The partial product through $n = 2\times10^{5}$ is $1.4650505520$ , short of the limit by $1.8\times10^{-6}$ , which is the right size: the logarithm of the tail is $\approx\sum_{n>N}1/(4n^{2}) \approx 1/(4N) = 1.25\times10^{-6}$ .

Step 4: the matching pole expansion. Differentiating $\log$ of the product termwise,

\frac{P'(z)}{P(z)} \;=\; \sum_{n=1}^{\infty}\frac{-1/n^{2}}{1 - z/n^{2}} \;=\; \sum_{n=1}^{\infty}\frac{1}{z-n^{2}},

a Mittag-Leffler expansion with simple poles at the squares, all of residue $1$ . The closed form gives the same thing another way: with $w = \sqrt z$ and $d/dz = (2w)^{-1}d/dw$ ,

\frac{P'(z)}{P(z)} \;=\; \frac{1}{2w}\,\frac{d}{dw}\log\frac{\sin\pi w}{\pi w} \;=\; \frac{\pi\cot\pi w - 1/w}{2w},

the cotangent expansion of Part C reappearing. At $z = -1/4$ , $w = i/2$ : $\pi\cot(i\pi/2) = -i\pi\coth(\pi/2)$ and $1/w = -2i$ , while $2w = i$ , so the right-hand side is $2 - \pi\coth(\pi/2)$ . The left-hand side is $-\sum_{n\ge1}4/(1+4n^{2})$ . Hence

\sum_{n=1}^{\infty}\frac{1}{1+4n^{2}} \;=\; \frac{\pi\coth(\pi/2) - 2}{4} \;=\; 0.3563442875 ,

and summing the series numerically gives $0.3563442875$ as well. The two checks test the product and the pole expansion against each other at the same point.

Application(Where these expansions are used)Original

Modal decomposition of a response function. A linear system's transfer function $\chi(\omega)$ is meromorphic, and its Mittag-Leffler expansion is the statement that the response is a sum of one simple resonance per pole, with weight the residue. In circuit and control theory this is the partial-fraction expansion that turns a transfer function into a sum of first-order sections; in scattering theory it is the resonance expansion of an amplitude, each pole a quasi-bound state with its width. The convergence terms $1/z_n$ are the non-resonant background that makes the sum of resonances converge, and they are not optional.
Sums over Matsubara frequencies. In finite-temperature field theory and in linear-response calculations one meets sums over the discrete frequencies $\omega_n = 2\pi n/\beta$ . The pole expansion of $\coth$ — the cotangent expansion of Part C along the imaginary axis — is exactly the kernel that converts such a sum into a contour integral, which is then evaluated by residues (Summing series with residues).
Spectral determinants and zeta regularisation. For an operator with eigenvalues $\lambda_n\to\infty$ , the Weierstrass product $\prod_n(1-\lambda/\lambda_n)e^{\cdots}$ is the entire function whose zeros are the spectrum: a regularised version of $\det(\lambda - A)$ . Which exponential factors are needed — the genus — is set by how fast the eigenvalues grow, and the entire factor $e^{g}$ is the ambiguity that the regularisation scheme has to fix. The one-dimensional model is $\sin\pi z$ , the spectral determinant of $-d^{2}/dx^{2}$ on an interval with Dirichlet ends.
Jost functions and bound states. In one-dimensional and radial scattering the Jost function is entire in the momentum variable, and its zeros are the bound states. Writing it as a Weierstrass product over those zeros, and its logarithmic derivative as a pole expansion, is the standard route to sum rules relating the bound-state energies to integrals of the phase shift.
Classical identities. Evaluating the cotangent expansion near $z = 0$ and matching Laurent coefficients gives $\sum_n n^{-2k}$ in closed form, and the sine product at $z = 1/2$ gives Wallis's product for $\pi$ . The number of zeros of such a product in a region is counted by the argument principle (Argument principle and Rouche's theorem).

Definition · advanced

Riemann surfaces

Adapted

Add to PDF

The domain on which a multivalued function becomes an honest single-valued analytic function, built by stacking cut copies of the plane and gluing them across the cuts at the branch points, with the surfaces of $z^{1/n}$ and $\log z$ worked out and parametrised by a single variable.

Prerequisites: Branch points and branch cuts, Analytic continuation, Complex numbers and elementary functions, Classification of isolated singularities, Laurent series, Connectedness and simple connectivity, Continuity and homeomorphism, Topological and metric spaces

NoteOriginal

What it means. A multivalued "function" is not a function: $z^{1/2}$ offers two values at each $z \ne 0$ , and $\log z$ offers infinitely many. Branch cuts deal with this by forbidding the paths that cause trouble, at the price of a barrier across which the chosen branch jumps. Riemann's idea is the opposite: keep every path and enlarge the domain instead.

Take as many copies of the cut plane as there are values, stack them, and glue the edge of each cut to the edge of the next in the order that continuation visits them. On the resulting surface each point carries not only a position $z$ but also a record of which value is in force there, so the function is genuinely single-valued, and genuinely analytic, with no jumps anywhere: the cut has been glued shut. The branch points are the hinges where the sheets meet, and they are the only places where the surface fails to look like an ordinary piece of plane.

The pay-off is that the awkward object becomes an ordinary one. For $z^{1/2}$ the whole surface turns out to be a single copy of the plane, coordinatised by $w$ with $z = w^{2}$ ; for $\log z$ it is again a plane, coordinatised by $w$ with $z = e^{w}$ . In that coordinate there is nothing multivalued left to discuss.

Definition(Sheets, gluing, and the surface of a multivalued function)AdaptedAdapted from Ahlfors (1979), Ch. 3, §4.3, pp. 97-98; Hassani (2013), §12.2.1, pp. 366-367

Let $f$ be multivalued on a region, with its branch points located as in Branch points and branch cuts. The construction is descriptive; Ahlfors introduces it this way deliberately and warns that it is for illustration, not for use inside proofs.

Choose cuts joining the branch points to each other or to $\infty$ , so that on the cut plane the values of $f$ separate into single-valued analytic branches.
Take one copy of the cut plane for each branch. These copies are the sheets. Points of different sheets may sit over the same $z$ ; they are distinguished by a label, so a point of the surface is a pair (position, label).
Each cut has two edges, an upper and a lower. Glue the edges in the pattern dictated by continuation: if crossing the cut at a given place carries the branch labelled $j$ into the branch labelled $k$ , attach that edge of sheet $j$ to the opposite edge of sheet $k$ . For $\ln z$ with the cut on the negative reals, Hassani attaches sheet $n-1$ to sheet $n$ by identifying the cut's upper edge on the former with its lower edge on the latter.

The space obtained is the Riemann surface of $f$ , and the map sending a point to its position $z$ is the projection $\pi$ . The cuts are not features of the surface: a different choice of cuts gives the same surface, and the cuts are introduced only to describe it. In the plane the gluing cannot be carried out without self-intersection; the surface is the idealised object in which no such crossing occurs.

A point of the surface over a branch point is a branch point of the surface. Following Arfken, Weber and Harris and Branch points and branch cuts, we call it of order $h$ when $h$ circuits of it are needed to restore the starting value, so that $h$ sheets meet there; the order is infinite when no number of circuits restores it. When the multivalued $f$ inverts a single-valued map $F$ , a region that $F$ carries one-to-one onto the whole plane with the cuts removed is a fundamental region of $F$ , and the sheets stand in one-to-one correspondence with the fundamental regions.

Definition(The surface as a set of germs)AdaptedAdapted from Ahlfors (1979), Ch. 8, §1.1-1.2, pp. 284-286; Ahlfors (1979), Ch. 8, §1.3, p. 288; Ahlfors (1979), Ch. 8, §1.4, p. 289

The descriptive construction can be replaced by one that needs no cuts and no pictures. Recall from Analytic continuation that a function element is a pair $(g, \Omega)$ with $g$ analytic on a region $\Omega$ , and that two elements are direct continuations of one another when they agree on a non-empty overlap. Two elements determine the same germ at a point $\zeta$ when they agree on some neighbourhood of $\zeta$ .

Let $\mathcal{G}$ be the set of all germs of analytic functions at all points of the plane, topologised so that the germs coming from one function element $(g,\Omega)$ , one at each point of $\Omega$ , form a neighbourhood of each of them. The projection $\pi$ sends a germ at $\zeta$ to $\zeta$ . A global analytic function is a connected component $\mathcal{G}_0$ of $\mathcal{G}$ : by construction it consists of exactly the germs obtainable from any one of them by a chain of direct continuations. Ahlfors calls $\mathcal{G}_0$ the Riemann surface of that global analytic function, and the function itself is defined on $\mathcal{G}_0$ by sending each germ to the value of the corresponding power series at its own centre.

Two things are immediate from this definition and are what make it useful.

$\pi$ is a local homeomorphism: restricted to the germs of one function element it is inverse to the map $\zeta\mapsto$ (germ at $\zeta$ ).
Continuation of $f$ along an arc $\gamma$ in the plane is the same thing as a lift of $\gamma$ : an arc $\tilde\gamma$ in $\mathcal{G}_0$ with $\pi\circ\tilde\gamma = \gamma$ .

Branch points are not germs and so are not yet points of $\mathcal{G}_0$ ; they are added separately, as the next statement describes.

Definition(What the surface does)AdaptedAdapted from Ahlfors (1979), Ch. 8, §1.4, Thm 1, p. 289; Ahlfors (1979), Ch. 8, §1.7, Branch Points, pp. 297-300; Ahlfors (1979), Ch. 3, §4.3, p. 98

Let $f$ be a global analytic function with Riemann surface $\mathcal{G}_0$ and projection $\pi$ .

Single-valuedness. $f$ is a single-valued function on $\mathcal{G}_0$ , and it is analytic there in the coordinate supplied by $\pi$ near any point that is not a branch point.
Unique lifting. Two continuations of $f$ along the same arc $\gamma$ either coincide at every parameter value or differ at every parameter value. A continuation is therefore determined by its initial germ.
Local structure at a branch point. Suppose $f$ can be continued along every arc of a punctured disc $0 < \lvert z - a\rvert < \rho$ , and let $h$ be the least positive integer such that $h$ circuits of $a$ return the initial germ to itself, if such an integer exists. Then in the variable $t$ with $z - a = t^{h}$ there is a single-valued analytic $F$ on $0 < \lvert t\rvert < \rho^{1/h}$ with $F(t) = f(a + t^{h})$ in the sense of continuation, and $F$ has a Laurent expansion there. Ahlfors calls $t$ the local uniformizing variable. When that expansion has only finitely many negative powers and $h > 1$ , the point is an algebraic singularity, $f$ tends to a definite limit (possibly $\infty$ ) as $z \to a$ along any arc, and the surface is completed by adding one branch point over $a$ .
It is a surface. After the branch points have been added in this way, every point of the completed space — branch points included — has a neighbourhood homeomorphic to a disc. The disc coordinate at an ordinary point is $z$ ; at a branch point of order $h$ it is $t = (z-a)^{1/h}$ .

Idea of the proofOriginal

The whole construction rests on one observation: continuation along a path is unique once the starting value is chosen, so a point of the plane plus a choice of value is enough information to say what happens next. Declare such a pair to be a point, and the multivalued function becomes a map on the set of pairs. Everything else is checking that the set of pairs is a reasonable space.

It is reasonable because locally there is no ambiguity at all. Away from the branch points a small disc in the plane carries exactly one value of $f$ per sheet, so near each pair the set of pairs looks like a disc, and the projection to the plane is a homeomorphism there. Only at a branch point do the sheets touch, and there the fix is to change coordinate: if $h$ turns are needed to close up, then $t = (z-a)^{1/h}$ makes one turn in $t$ correspond to $h$ turns in $z$ , and in $t$ the function is single-valued again. That is why the branch point, too, has a disc around it.

For the two basic examples one can do better than "looks locally like a disc". Solving $w^{2} = z$ for $z$ rather than for $w$ gives $z = w^{2}$ , which is single-valued; so the set of pairs $(z,w)$ is nothing but the $w$ -plane, and the two-valued $z^{1/2}$ has become the identity map $w \mapsto w$ . The same trick on $e^{w} = z$ makes the surface of $\log$ the $w$ -plane. Making a multivalued relation single-valued by parametrising it is what "uniformizing" means.

Proof(The surfaces of the power, the logarithm, and their gluing)AdaptedAdapted from Ahlfors (1979), Ch. 3, §4.3, pp. 97-98, and Ch. 8, §1.4, p. 290; Hassani (2013), §12.2.1, pp. 366-368

Step 1: the surface of $z^{1/n}$ , as a set of pairs. Fix an integer $n \ge 2$ and set

S_{n} \;=\; \bigl\{(z,w) \in \mathbb{C}^{2} \;:\; w^{n} = z \bigr\}, \qquad \pi(z,w) = z .

The map $\Phi(w) = (w^{n}, w)$ sends $\mathbb{C}$ into $S_n$ ; it is injective because the second coordinate is recovered from the pair, and surjective because every point of $S_n$ is $(w^n,w)$ for its own $w$ . Both $\Phi$ and its inverse $(z,w)\mapsto w$ are continuous, so $S_n$ is homeomorphic to the plane. On it the function "an $n$ -th root of $z$ " is the single-valued map $(z,w)\mapsto w$ , which in the coordinate $w$ is the identity, hence analytic. This is item 1 of the statement, with the coordinate of item 3 made global: here $t = w$ and $z = t^{n}$ everywhere, not just near the branch point.

The projection is $n$ -to-one over every $z \ne 0$ , since $w^{n} = z$ has $n$ distinct roots there (Complex numbers and elementary functions), and one-to-one over $z = 0$ . So over the finite plane $S_n$ has $n$ sheets and exactly one branch point, the point with $w = 0$ . Over the extended plane there is a second: both Ahlfors (p. 98) and Hassani (Example 12.2.2) count $\infty$ as a branch point of $z^{1/n}$ as well, and it is of the same order, since the substitution $z \mapsto 1/z$ turns $z^{1/n}$ into $z^{-1/n}$ , whose branch point at the origin again needs $n$ circuits.

Step 2: the gluing, recovered. Cut the $z$ -plane along the positive reals and let $\theta = \arg z$ run over $0 < \theta < 2\pi$ . The $n$ branches are

f_k(z) = r^{1/n}\exp\!\left[i\,\frac{\theta + 2k\pi}{n}\right], \qquad k = 0, 1, \dots, n-1, \quad r = \lvert z\rvert ,

each analytic on the cut plane, and they are exactly the $n$ values of $w$ . (Example 12.2.2 of Hassani prints this exponent as $i(\theta + 2k\pi/n)$ , with the $1/n$ dividing only the second term; as printed those are not $n$ -th roots of $z$ , and the sheet count and pasting given on the same page need the form above, so the printed exponent is a typesetting slip and not a disagreement with what follows.) Their images are the sectors $2k\pi/n < \arg w < 2(k+1)\pi/n$ of the $w$ -plane, so each sector is a fundamental region and each branch is one sheet. Approaching the cut from below, $\theta \to 2\pi$ , gives $f_k \to r^{1/n}e^{i(2\pi + 2k\pi)/n} = f_{k+1}(r)$ read from above, where $f_{k+1}$ means $f_0$ when $k = n-1$ . The lower edge of sheet $k$ therefore joins the upper edge of sheet $k+1$ , cyclically, the last to the first: Ahlfors's cyclic attachment and Hassani's pasting of the lower edge of $R_{n-1}$ to the upper edge of $R_0$ . Going once round the origin advances $k$ by one, so $n$ circuits are needed to return: the branch point has order $n$ in the convention of Branch points and branch cuts.

Step 3: the surface of $\log z$ . Set

S_{\log} \;=\; \bigl\{(z,w) \in \mathbb{C}^{2} \;:\; e^{w} = z\bigr\},\qquad \pi(z,w) = z .

As before $w \mapsto (e^{w}, w)$ is a homeomorphism of $\mathbb{C}$ onto $S_{\log}$ , so this surface is also a plane, with $\log$ the single-valued map $(z,w)\mapsto w$ . Two differences from Step 2 are worth naming. First, $e^{w}$ is never $0$ , so there is no point of $S_{\log}$ over $z = 0$ : the origin is missing from the surface, not a branch point of it. Second, $\pi$ is now infinitely many to one, since $e^{w} = e^{w + 2\pi i m}$ for every integer $m$ , and no number of circuits of the origin returns the starting value. The sheets, in Hassani's description, are the horizontal strips $(2n-1)\pi < \operatorname{Im} w < (2n+1)\pi$ , which correspond to the plane cut along the negative reals; continuing counterclockwise across the cut, from just above the negative reals to just below them, carries sheet $n-1$ into sheet $n$ , so the sheets form Ahlfors's endless screw rather than a cycle.

Step 4: these agree with the germ definition. For the logarithm this is Ahlfors's argument on p. 290. All function elements $(g,\Omega)$ with $e^{g(\zeta)} = \zeta$ on $\Omega$ form a single global analytic function: take two of them and fix a point $\zeta_1$ in the first domain and $\zeta_2$ in the second. Since a logarithm is defined at each, neither is the origin, so some arc $\gamma : [0,1] \to \mathbb{C}\setminus\{0\}$ runs from $\zeta_1$ to $\zeta_2$ . Put $\varphi(s) = g_1(\zeta_1) + \int_{0}^{s}\gamma'(u)/\gamma(u)\,du$ . Then $e^{\varphi(s)} = \gamma(s)$ for every $s$ : the quotient $\gamma e^{-\varphi}$ is constant, since its derivative vanishes identically, and it takes the value $1$ at $s = 0$ . Consequently the germs of $\varphi$ along $\gamma$ continue the first element to a germ over the second point, and the two differ at most by a multiple of $2\pi i$ , which is removed by continuing further around the origin the right number of times. Hence the germs of $\log$ form one connected component of $\mathcal{G}$ , and sending a germ at $\zeta$ to the pair $(\zeta, w)$ , where $w$ is its value at $\zeta$ , identifies that component with $S_{\log}$ : the germ determines the pair, and the pair determines the germ, because the branch of $\log$ near $\zeta$ taking a prescribed value at $\zeta$ is unique. The same argument with $g(\zeta)^{n} = \zeta$ identifies the germs of $z^{1/n}$ with $S_{n}$ minus its branch point, which Step 1's completion restores.

Step 5: the lift of a loop, and why it detects the branch point. Let $\gamma(s) = \rho e^{is}$ , $0 \le s \le 2\pi$ . In $S_n$ the lifts of $\gamma$ are the arcs $s\mapsto(\rho e^{is}, \rho^{1/n}e^{i(s + 2k\pi)/n})$ , since these satisfy $w^{n} = z$ and project correctly, and by item 2 of the statement there are no others once the initial point is fixed. The lift starting at $w = \rho^{1/n}$ ends at $\rho^{1/n}e^{2\pi i/n} \ne \rho^{1/n}$ for $n \ge 2$ : the loop does not close on the surface, which is the precise sense in which the sheet has changed. In $S_{\log}$ the lift starting at $w = \ln\rho$ ends at $\ln\rho + 2\pi i$ , and iterating never returns. $\square$

Remark(Where the hypotheses are used, and two conventions)Original

Item 3 needs continuation along every arc of the punctured disc. The integer $h$ is defined by following the germ around the puncture; if continuation fails along some arc — if, say, there is a second branch point inside the disc, or a natural boundary — there is no single $h$ and no local uniformizing variable. This is why the first step of any construction is to find all the branch points, $\infty$ included, and then work on discs small enough to isolate one of them.
Item 3 also needs the least $h$ to exist. For $\log z$ it does not: no number of circuits restores the value, the branch point has infinite order, and the substitution $z = t^{h}$ is unavailable. The surface still exists — Step 3 built it — but it has no point over the origin, and the origin is a logarithmic branch point rather than an algebraic one.
The cuts are scaffolding. Nothing in the germ definition mentions them, and Ahlfors states explicitly (p. 98) that the cut may be replaced by any simple arc from the branch point to $\infty$ without changing the surface. What the cuts do is make the sheets visible. A consequence worth keeping: a statement about "the jump across the cut" is a statement about a chosen branch, not about the surface.
A branch point need not join all the sheets. Ahlfors's convention is that a point joining $h$ sheets has order $h - 1$ ; Arfken, Weber and Harris count instead the circuits needed to restore the value, which for $z^{1/2}$ gives $2$ where Ahlfors gives $1$ . The two differ by one throughout. We follow Arfken, Weber and Harris here, as Branch points and branch cuts does, and the worked example below says which number is meant each time it is used.
Single-valuedness is bought, not free. The surface makes $f$ single-valued by enlarging the domain, so a formula valid on the surface says nothing by itself about which value to use at a given $z$ in the plane. Picking a value is picking a sheet, and that is exactly what a branch and its cut do.

Method(Constructing the surface of a given multivalued function)Original

Find the branch points, including the point at infinity, by the loop test of Branch points and branch cuts. For a function built from factors, test a small loop around each candidate and record how each factor's argument changes.
Find the order of each. Continue once around; count how many circuits restore the starting value. A finite count $h$ means $h$ sheets meet there; no count means a logarithmic branch point of infinite order.
Test loops around groups of branch points. A loop enclosing several may return to the starting sheet even though each individual loop does not; this decides whether a cut may be run between two finite branch points instead of out to infinity, and whether $\infty$ is a branch point.
Choose cuts so that the complement is simply connected, or at least carries no loop around a single unmatched branch point, and list the branches on the cut plane.
Glue. For each edge of each cut, continue a branch across it and read off which branch it becomes; attach that edge to the opposite edge of the corresponding sheet.
Check the count. The number of sheets must equal the number of values of $f$ at a generic point, and the cyclic order of the gluing at a branch point of order $h$ must close after $h$ steps.
Look for a global parameter. If the defining relation can be solved for $z$ in terms of the value, as in $z = w^{n}$ or $z = e^{w}$ , that solution parametrises the whole surface and ends the problem.

Example(The two-sheeted surface of

(z^{2}-1)^{1/2}

, with continuations checked)AdaptedAdapted from Hassani (2013), §12.2.1, Example 12.2.3, p. 368; Arfken, Weber & Harris (2013), §11.6, Example 11.6.4, pp. 502-503

Let $f(z) = (z^{2}-1)^{1/2} = (z-1)^{1/2}(z+1)^{1/2}$ . Write $z - 1 = r_{1}e^{i\theta_{1}}$ and $z+1 = r_{2}e^{i\theta_{2}}$ , so that one value of $f$ is $\sqrt{r_{1}r_{2}}\,e^{i(\theta_{1}+\theta_{2})/2}$ and the other is its negative.

Branch points. A loop around $z = 1$ alone sends $\theta_{1}\mapsto\theta_{1}+2\pi$ and returns $\theta_{2}$ , so $f \mapsto f e^{i\pi} = -f$ : $z = 1$ is a branch point, and a second circuit restores $f$ , so its order is $2$ . The same holds at $z = -1$ . At $\infty$ : a loop enclosing both sends $\theta_{1}+\theta_{2}\mapsto \theta_{1}+\theta_{2}+4\pi$ , so $f\mapsto fe^{2\pi i} = f$ . Hence $\infty$ is not a branch point, and the cut may be taken as the finite segment $[-1,1]$ . Two values at a generic point means two sheets, joined along that segment, each sheet being a copy of $\mathbb{C}\setminus[-1,1]$ .

Numerical check of the gluing. Take the circle $\lvert z - 1\rvert = 3/2$ , which starts and ends at $z = 5/2$ and encloses $z=1$ but not $z=-1$ (the distance from the centre to $-1$ is $2 > 3/2$ ). Start from the value

f(5/2) \;=\; \sqrt{(5/2)^{2}-1} \;=\; \sqrt{5.25} \;=\; 2.2912878475 .

Following the circle in $4\times10^{5}$ steps and choosing at each step whichever square root is nearer the previous value — which is continuation, by uniqueness of the lift — the value on return is

-2.2912878475 \quad (\text{imaginary part } 4\times10^{-16}),

the other sheet. A second circuit brings it back. Repeating on the circle $\lvert z\rvert = 3$ , which encloses both branch points, starting from $f(3) = \sqrt{8} = 2.8284271247$ , the value on return is $2.8284271247$ to ten digits: the loop closes on the surface, confirming that $\infty$ is not a branch point.

The surface named. Solve for $z$ : the surface is $\{(z,w) : w^{2} = z^{2}-1\}$ , and setting $z = \cosh\tau$ , $w = \sinh\tau$ satisfies the relation identically. The map $\tau\mapsto(\cosh\tau,\sinh\tau)$ is onto the surface and identifies points differing by $2\pi i$ , so the surface is the cylinder $\mathbb{C}/2\pi i\mathbb{Z}$ — equivalently, via $u = e^{\tau}$ , the punctured plane. That is a uniformization in the sense of the Idea: on the $\tau$ -cylinder both $z$ and $w$ are single-valued analytic functions, and the relation $w^{2} = z^{2}-1$ holds with no choice of branch to make. The two sheets are the two halves $\operatorname{Re}\tau > 0$ and $\operatorname{Re}\tau< 0$ , and the segment $[-1,1]$ is the image of the line $\operatorname{Re}\tau = 0$ , which is where they meet.

The logarithm, for contrast. On $S_{\log}$ start at the point over $z = 1$ with $w = 0$ , wind twice counterclockwise around the origin, and finish along a path ending at $z = -2$ with a further half turn. The lift has $w = \ln\lvert z\rvert + i\arg z$ with $\arg z$ tracked continuously, so the total argument gained is $2\cdot2\pi + \pi = 5\pi$ and the value is

w \;=\; \ln 2 + 5\pi i \;=\; 0.6931471806 + 15.7079632679\,i .

Running the same path with one loop instead of two gains $2\pi + \pi = 3\pi$ and gives $\ln 2 + 3\pi i = 0.6931471806 + 9.4247779608\,i$ ; omitting the loops altogether and going straight to $z = -2$ gains only $\pi$ and gives $\ln 2 + \pi i$ . On the surface these are three different points over the same $z = -2$ , and no number of further loops ever identifies them.

Application(Where the sheets matter)Original

Resonances live on the second sheet. A scattering amplitude as a function of complex energy has a square-root branch point at each two-body threshold, because the channel momentum is $k = (2\mu(E - E_{\text{th}}))^{1/2}$ . The surface is two-sheeted at a single threshold, and a resonance is a pole not on the physical sheet — where it would violate causality — but on the second sheet, reached by passing through the cut above the threshold. The pole position $E_{0} - i\Gamma/2$ there gives the mass and width. Which sheet a pole sits on is not a technicality: it is the difference between a bound state and a decaying one.
Radiation conditions in wave problems. In a waveguide, a stratified medium, or a plasma, the longitudinal wavenumber $k_{z} = (k^{2}-k_{\perp}^{2})^{1/2}$ is two-valued, and the two sheets are "outgoing" and "incoming", or "decaying" and "growing", in the evanescent range. Imposing a radiation condition is choosing the sheet, and a numerical code that lets $k_z$ wander across the cut silently switches to the unphysical solution.
Periods and the pendulum. When $w^{2}$ equals a cubic or quartic in $z$ with distinct roots, the surface, once completed at infinity, is a torus rather than a plane, and the two independent loops on it carry two independent periods. The exact pendulum qualifies: energy conservation makes its angular velocity the square root of a trigonometric polynomial in the angle, which a half-angle substitution turns into such a curve, and the Jacobi elliptic functions are the uniformizing parameter. One of the two periods is then the period of the oscillation.
Analytic continuation of transforms and Green's functions. Continuing a response function or a Feynman integral past its cut, to look for poles or to deform a contour, is motion on a Riemann surface, and the sheets record which thresholds have been crossed. Dispersion relations are statements on the physical sheet; the singularity structure on the others is what controls the long-time or short-distance behaviour.
Conformal mapping of cut domains. A flow or potential problem on a slit domain is often solved by mapping the slit plane to a half-plane with a square root (Conformal mappings); the two sheets of that square root are the two sides of the slit, so the Riemann-surface picture says exactly which physical region each part of the image represents.