Analytic functions and Cauchy's theorem

1. Addition theorem. Both series converge absolutely, so their product may be summed by diagonals (absolute convergence):

the middle step being the binomial theorem.

2. No zeros. Put $b = -a$ in part 1: $e^a e^{-a} = e^0 = 1$. A product equal to $1$ has no zero factor.

3. Modulus. The coefficients $1/n!$ are real, so conjugating the series termwise gives $\overline{e^z} = e^{\bar z}$. Hence $\lvert e^{iy}\rvert^2 = e^{iy}\overline{e^{iy}} = e^{iy}e^{-iy} = 1$ for real $y$, and by part 1, $\lvert e^{x+iy}\rvert = \lvert e^x\rvert\,\lvert e^{iy}\rvert = e^x$, using that $e^x > 0$ for real $x$: for $x \ge 0$ every term of the series is non-negative and the first is $1$, and for $x < 0$ we have $e^x = 1/e^{-x} > 0$ by part 2.

4. Periodicity. A period exists. Write $C(y) = \tfrac12(e^{iy} + e^{-iy})$ and $S(y) = \tfrac{1}{2i}(e^{iy} - e^{-iy})$ for real $y$; from the series these are the real functions $\cos$ and $\sin$, with $C' = -S$, $S' = C$, $C(0)=1$, $S(0)=0$, and $C^2 + S^2 = 1$ because $C^2 + S^2 = e^{iy}e^{-iy} = 1$ by parts 1 and 2. In particular $\lvert C\rvert \le 1$ and $\lvert S \rvert \le 1$. Four successive integrations from $0$ now trap $C$ and $S$ between consecutive partial sums of their series. From $C \le 1$ and $S(0)=0$ we get $S(y) \le y$ for $y \ge 0$; integrating $C' = -S \ge -y$ gives $C(y) \ge 1 - y^2/2$; integrating $S' = C \ge 1 - y^2/2$ gives $S(y) \ge y - y^3/6$; integrating $C' = -S \le -y + y^3/6$ gives

At $y = \sqrt3$ the right-hand side is $1 - \tfrac32 + \tfrac38 = -\tfrac18 < 0$. Since $C$ is continuous with $C(0) = 1$, it has a zero in $(0, \sqrt 3)$; let $y_0$ be the smallest one. From $C(y_0) = 0$ and $C^2 + S^2 = 1$ we get $S(y_0)^2 = 1$; and $S(y_0) > 0$, because $S(y) \ge y - y^3/6 > 0$ for $0 < y < \sqrt 6$ and $y_0 < \sqrt 3 < \sqrt 6$. Hence $S(y_0) = 1$, so

the number $4y_0$ is a period of $y \mapsto e^{iy}$.

It is the smallest. For $0 < y < y_0$ we have $C(y) > 0$ (by minimality of $y_0$ and continuity, since $C(0)=1$), so $S' = C > 0$ there and $S$ increases strictly from $S(0) = 0$. Hence $0 < S(y) < S(y_0) = 1$ on $(0,y_0)$, and therefore $C(y)^2 = 1 - S(y)^2 \in (0,1)$. So for $0 < y < y_0$ the number $e^{iy} = C(y) + iS(y)$ has both parts nonzero and is not one of $1, -1, i, -i$; consequently $e^{4iy} \ne 1$, because $w^4 = 1$ forces $w \in \{1,-1,i,-i\}$. No positive period is therefore smaller than $4y_0$.

All periods are multiples. Let $\omega > 0$ be a period and $\omega_0 = 4y_0$. Choose the integer $n$ with $n\omega_0 \le \omega < (n+1)\omega_0$. If $\omega \ne n\omega_0$ then $\omega - n\omega_0$ is a positive period smaller than $\omega_0$, which we have excluded. Finally, $e^{z+c} = e^z$ for all $z$ forces $e^c = 1$, so $\lvert e^c \rvert = 1$ gives $\operatorname{Re} c = 0$ by part 3, and then $c = i\omega$ with $\omega$ a period of $y \mapsto e^{iy}$. Writing $2\pi = 4y_0$ gives the displayed statement.

5. Surjectivity. Let $w \ne 0$ and seek $z = x+iy$. By part 3, $e^{x+iy} = w$ splits into $e^x = \lvert w\rvert$ and $e^{iy} = w/\lvert w\rvert$. The first has the unique real solution $x = \log\lvert w\rvert$, since the real exponential is a strictly increasing bijection of $\mathbb{R}$ onto $(0,\infty)$. For the second, the map $y \mapsto e^{iy}$ sends $[0, 2\pi)$ onto the unit circle: it is continuous; part 4 showed that $S > 0$ on $(0,y_0)$, so $C' = -S < 0$ there and $C$ decreases strictly from $C(0) = 1$ to $C(y_0) = 0$, which together with $C^2 + S^2 = 1$ and $S \ge 0$ traces the first quadrant of the circle exactly once, and the four reflections $e^{i(y + k\pi/2)} = i^k e^{iy}$ carry this arc onto the remaining three quadrants. Two solutions of $e^{iy} = w/\lvert w\rvert$ differ by a period, by part 4, so the solution set of $e^z = w$ is a single coset of $2\pi i\,\mathbb{Z}$. $\square$

De Moivre's theorem is the addition theorem in polar clothing: $\cos\theta + i\sin\theta = e^{i\theta}$ by Euler's formula, and $(e^{i\theta})^n = e^{in\theta}$ for $n \ge 0$ by part 1 of the statement above and induction, while for $n < 0$ one uses $e^{-i\theta} = 1/e^{i\theta}$ from part 2.

For the roots, write $z = r e^{i\vartheta}$ with $r > 0$; this costs nothing, because $z^n = a \ne 0$ forces $z \ne 0$, and every nonzero $z$ has such a form by the definition of polar form. Then $z^n = r^n e^{in\vartheta}$, and $z^n = a = \rho e^{i\varphi}$ holds if and only if the moduli agree and the arguments agree modulo a period:

• $r^n = \rho$, which by strict monotonicity of $r \mapsto r^n$ on $(0,\infty)$ fixes $r = \rho^{1/n}$ uniquely;
• $e^{in\vartheta} = e^{i\varphi}$, which by part 4 means $n\vartheta - \varphi \in 2\pi\mathbb{Z}$, that is $\vartheta = (\varphi + 2k\pi)/n$ for some $k \in \mathbb{Z}$.

Two integers $k$ and $k'$ give the same $z$ exactly when $(k - k')/n \in \mathbb{Z}$, so $k = 0, 1, \dots, n-1$ list the solutions once each: there are exactly $n$. Their moduli are all $\rho^{1/n}$ and consecutive arguments differ by $2\pi/n$, which is the description of a regular $n$-gon. Dividing two solutions gives a solution of $z^n = 1$, which proves the last sentence. $\square$

Part 1 (necessity). Suppose $f'(z_0)$ exists. Restrict $h$ to real values: then $y$ is held at $y_0$ and the difference quotient becomes the difference quotient of $x \mapsto f(x + iy_0)$. Its limit exists and equals $f'(z_0)$, so $u_x$ and $v_x$ exist at $z_0$ and

Now restrict $h$ to purely imaginary values $h = ik$ with $k$ real. Since $1/(ik) = -i/k$,

so $u_y$ and $v_y$ exist at $z_0$ as well. The two expressions for $f'(z_0)$ are the same complex number; equating real parts gives $u_x = v_y$ and imaginary parts gives $v_x = -u_y$. (The combined form is $f_x = -if_y$, which on substituting $f = u+iv$ reproduces the pair.)

Part 2 (sufficiency). Fix $z_0 \in \Omega$ and let $h + ik$ be a small complex increment, $h, k$ real. Because $u$ has continuous first partials on the open set $\Omega$, it is differentiable there in the sense of the calculus of several variables:

with the partials evaluated at $z_0$ and $\varepsilon_1(h,k)/\lvert h + ik\rvert \to 0$ as $h + ik \to 0$; and likewise

This is the one place where continuity of the partials does essential work; it is used once more, trivially, at the end of the part, to see that $f' = u_x + iv_x$ is continuous. Adding $i$ times the second to the first,

Now apply the hypothesis: by the Cauchy-Riemann equations at $z_0$, $u_y + iv_y = -v_x + iu_x = i(u_x + i v_x)$. Hence the two linear terms combine,

and dividing by $h + ik \ne 0$,

The last term tends to $0$ as $h + ik \to 0$, because $\lvert \varepsilon\rvert/\lvert h+ik\rvert \le \lvert\varepsilon_1\rvert/\lvert h+ik\rvert + \lvert \varepsilon_2\rvert/\lvert h+ik\rvert \to 0$. So the limit exists, independently of the direction of approach, and $f'(z_0) = u_x + iv_x$. As $z_0 \in \Omega$ was arbitrary, $f$ is analytic on $\Omega$, and $f' = u_x + iv_x$ is continuous because $u_x$ and $v_x$ are.

Part 3 (harmonicity). Differentiate $u_x = v_y$ with respect to $x$ and $u_y = -v_x$ with respect to $y$:

The second partial derivatives of $v$ are continuous by hypothesis, so the mixed partials agree, $v_{yx} = v_{xy}$. Adding the two displayed equations therefore gives $u_{xx} + u_{yy} = 0$. For $v$, note that $-if = v - iu$ is differentiable wherever $f$ is, with real part $v$ and imaginary part $-u$; applying the result just proved to $-if$ gives $v_{xx} + v_{yy} = 0$. (Equivalently, differentiate $u_x = v_y$ with respect to $y$ and $u_y = -v_x$ with respect to $x$ and subtract.)

Uniqueness of the conjugate. If $v$ and $\tilde v$ both satisfy the Cauchy-Riemann equations with the same $u$, then $w = v - \tilde v$ has $w_x = w_y = 0$ on $\Omega$, so $w$ is constant on each connected component: constant on $\Omega$ if $\Omega$ is connected. $\square$