Cauchy’s Theorem

• Prove a more general statement: if ${\left\lvert {f(z)} \right\rvert} \leq M{\left\lvert {z} \right\rvert}^n$, then $f$ is a polynomial of degree at most $n$.

• Since $f$ is entire, it is analytic everywhere, so $f(z) = \sum_{k\geq 0}c_k z^k$ where $c_k = f^{(k)}(0)/n!$ is given by the coefficient of its Taylor expansion about $z=0$.

• Applying Cauchy’s estimate, on a circle of radius $R$, $$\begin{align*} {\left\lvert {f^{(k)}(0)} \right\rvert} \leq { \sup_{\gamma}{\left\lvert {f(z)} \right\rvert} n! \over R^k} \leq {M{\left\lvert {z} \right\rvert}^n n! \over R^k} = {M R^n n! \over R^k} .\end{align*}$$

• So for $k \geq n+1$, this goes to zero as $R\to \infty$, so ${\left\lvert {f^{k}(0)} \right\rvert} = 0$ for all such $k$.

• But then $f$ is a power series annihilated by taking $n+1$ derivatives, so it is a polynomial of degree at most $n$.

$$\begin{align*} \int_\gamma z^n\,dz= \int_0^{2\pi} R^n e^{itn} \cdot iRe^{it} \,dt = R^{n+1} \int_0^{2\pi} e^{i(t+1)n}\,dt = { i R^{n+1} \over i(n+1) } \delta_{n+1 = 0} .\end{align*}$$

About a point $a$ and $R<{\left\lvert {a} \right\rvert}$, $$\begin{align*} \int_{{\left\lvert {z-a} \right\rvert} = R} z^n\,dz &= \int_0^{2\pi} (a + re^{it})^n \cdot ire^{it}\,dt\\ &= \int_0^{2\pi} \sum_{1\leq k\leq n} {n \choose k} a_k R^{n-k+1} e^{it(n-k)} \cdot ire^{it}\,dt\\ &= i \int_0^{2\pi} \sum_{1\leq k\leq n} {n \choose k} a_k R^{n-k+1} e^{it(n-k+1)} \,dt\\ &= i \sum_{1\leq k\leq n} {n \choose k} a_k R^{n-k+1} \int_0^{2\pi} e^{it(n-k+1)} \,dt\\ &= i \sum_{1\leq k\leq n} {n \choose k} a_k R^{n-k+1} \cdot 0 \\ &= 0 ,\end{align*}$$ provided $n\neq 0$, in which case $\int_\gamma \,dz= 2\pi$.

For the third computation, this follows from partial fraction decomposition.

Note that $G_z(\xi) \coloneqq{f(\xi) \over \xi - z}$ has a pole of order one at $\xi = z$ and also a pole at $\xi = \infty$. If $z\in \Omega_1$, then $\gamma$ encloses just the pole $\xi = z$, so apply the residue theorem: $$\begin{align*} F(z) &\coloneqq{1\over 2\pi i}\oint_\gamma {f(\xi) \over \xi - z}\,d\xi\\ &= {1\over 2\pi i}\oint_\gamma G_z(\xi) \,d\xi\\ &= \mathop{\mathrm{Res}}_{\xi = z} G_z(\xi) \\ &= \lim_{\xi\to z} (\xi - z) G_z(\xi) \\ &= \lim_{\xi\to z} (\xi - z) {f(\xi) \over \xi-z} \\ &= \lim_{\xi\to z} f(\xi) \\ &= f(z) .\end{align*}$$

Now if $z\in \Omega_2$, then $\gamma$ encloses both $\xi=z, \infty$, and is oriented negatively,so $$\begin{align*} F(z) &= {1\over 2\pi i} \oint_\gamma G_z(\xi) \,d\xi\\ &= -\qty{\mathop{\mathrm{Res}}_{\xi = z} G_z(\xi) + \mathop{\mathrm{Res}}_{\xi = \infty} G_z(\xi)}\\ &= -\qty{f(z) + \mathop{\mathrm{Res}}_{\xi = \infty} G_z(\xi)}\\ ,\end{align*}$$ where the last line proceeds by the same calculation as above. It remains to compute the unknown residue. Residues at $\xi = \infty$ are computed as residues at $\xi =0$, and the change of variables $G_z(\xi)\,d\xi\mapsto G_z(w) \,dw$ for $w\coloneqq 1/\xi$ yields $G_z(\xi)\,d\xi\to G_z\qty{1\over \xi}(-1/\xi^2)\,d\xi$. Thus $$\begin{align*} \mathop{\mathrm{Res}}_{\xi=\infty} G_z(\xi) &= -\mathop{\mathrm{Res}}_{\xi=0} G_z\qty{\xi^{-1}}\xi^{-2} \\ &= - \mathop{\mathrm{Res}}_{\xi=0} {f(\xi^{-1}) \over \xi^2(\xi^{-1}- z) } \\ &= - \mathop{\mathrm{Res}}_{\xi=0} {f(\xi^{-1}) \over \xi(1 - z\xi) } \\ &= -\lim_{\xi \to 0} {f(\xi^{-1}) \over 1-z\xi} \\ &= -\lim_{\xi \to 0}f(\xi ^{-1}) \\ &= -\lim_{\xi\to\infty} f(\xi) \\ &= -A .\end{align*}$$ So combining this yields $$\begin{align*} F(z) = -\qty{f(z) - A} = -f(z) + A .\end{align*}$$

Green’s theorem: if $\Omega$ is a domain with positively oriented boundary with $u, v$ continuously differentiable in $\overline{\Omega}$, then $$\begin{align*} \int_{{{\partial}}\Omega} u\,dx+ v\,dy= \iint_{\Omega}\qty{v_x - u_y}\,dx\,dy .\end{align*}$$ Now use that if $f = u+iv$ is analytic in a region, it satisfies Cauchy-Riemann: $$\begin{align*} u_x = v_y \qquad u_y = -v_x .\end{align*}$$

Now integrating $f$: $$\begin{align*} \oint_{{{\partial}}\Omega} f(z) \,dz &= \oint_{{{\partial}}\Omega} (u+iv)(\,dx+ i\,dy)\\ &= \oint_{{{\partial}}\Omega} \qty{u\,dx- v\,dy} + i\oint_{{{\partial}}\Omega} \qty{v\,dx+ u\,dy} \\ &= \iint_\Omega\qty{v_x + u_y}\,dx\,dy+ \iint_\Omega\qty{u_x - v_y}\,dx\,dy\\ &= \iint_\Omega\qty{v_x -v_x }\,dx\,dy+ \iint_\Omega\qty{u_x - u_x}\,dx\,dy\\ &= 0 .\end{align*}$$

• By Cauchy’s integral formula, $\int_{S^1} f = 2\pi i$
• If $p_j$ is any polynomial, then $p_j$ is holomorphic in ${\mathbb{D}}$, so $\int_{S^1} p_j = 0$.
• Contradiction: compact sets in ${\mathbf{C}}$ are bounded, so $$\begin{align*} {\left\lvert {\int f - \int p_j} \right\rvert} &\leq \int {\left\lvert {p_j - f} \right\rvert} \\ &\leq \int {\left\lVert {p_j - f} \right\rVert}_\infty \\ &= {\left\lVert {p_j - f} \right\rVert}_\infty \int_{S^1} 1 \,dz \\ &= {\left\lVert {p_j-f} \right\rVert}_\infty \cdot 2\pi \\ &\to 0 \end{align*}$$ which forces $\int f = \int p_j = 0$.

Let $R> 10$, then by Cauchy: $$\begin{align*} 2\pi {\left\lvert {f'(z)} \right\rvert} &\leq \oint_{{\left\lvert {\xi} \right\rvert} = R} { {\left\lvert { f(\xi)} \right\rvert} \over {\left\lvert {\xi} \right\rvert}^2 } \,d\xi\\ &\leq \oint_{{\left\lvert {\xi} \right\rvert} = R} R^{-2} {\left\lvert {\xi} \right\rvert}^{1\over 2} \,d\xi\\ &= R^{-{3\over 2}} \cdot 2\pi R \\ &\sim R^{-{1\over 2}} \\ &\overset{R\to\infty}\longrightarrow 0 .\end{align*}$$

Toward applying Morera, let $T \subseteq {\mathbf{C}}\setminus\gamma$ be a triangle, so that $z\in T$ and $w\in \gamma$ implies $z-w\neq 0$. Then $$\begin{align*} \oint_T f(z) \,dz &= \oint_T \int_\gamma {g(w)\over w-z}\,dw\,dz\\ &= \int_\gamma \oint_T {g(w)\over w-z}\,dz\,dw\\ &= \int_\gamma g(w) \qty{ \oint_T {1 \over w-z}\,dz} \,dw\\ &= \int_\gamma g(w) \cdot 0 \,dw\\ &= 0 ,\end{align*}$$ where the exchange of integrals is justified by compactness of $\gamma, T$, and the inner integral vanishes because for a fixed $w\in \gamma$, the function $z\mapsto {1\over w-z}$ has a simple pole at $w$, and so is holomorphic in $\gamma^c$ and vanishes by Goursat.

Just reproducing the proof of holomorphicity in the Schwarz reflection theorem.

• Note $f$ is continuous on ${\mathbf{C}}$ since analytic implies continuous ($f$ equals its power series, where the partials sums uniformly converge to it, and uniform limit of continuous is continuous).
• Strategy: take $D$ a disc centered at a point $x\in {\mathbf{R}}$, show $f$ is holomorphic in $D$ by Morera’s theorem.
• Let $\Delta \subset D$ be a triangle in $D$.
• Case 1: If $\Delta \cap{\mathbf{R}}= 0$, then $f$ is holomorphic on $\Delta$ and $\int_\Delta f = 0$.
• Case 2: one side or vertex of $\Delta$ intersects ${\mathbf{R}}$, and wlog the rest of $\Delta$ is in ${\mathbb{H}}^+$.
  • Then let $\Delta_{\varepsilon}$ be the perturbation $\Delta + i{\varepsilon}= \left\{{z+ i{\varepsilon}{~\mathrel{\Big\vert}~}z\in \Delta}\right\}$; then $\Delta_{\varepsilon}\cap{\mathbf{R}}= 0$ and $\int_{\Delta_{\varepsilon}} f = 0$.
  • Now let ${\varepsilon}\to 0$ and conclude by continuity of $f$ (???)
    • We want $$\begin{align*} \int_{\Delta_{\varepsilon}} f = \int_a^b f(\gamma_{\varepsilon}(t)) \gamma_{\varepsilon}'(t)\,dt \overset{{\varepsilon}\to 0}\to \int_a^b f(\gamma(t)) \gamma_{\varepsilon}'(t)\,dt =\int_{\Delta} f \end{align*}$$ where $\gamma_{\varepsilon}, \gamma$ are curves parametrizing $\Delta_{\varepsilon}, \Delta$ respectively.
    • Since $\gamma, \gamma_{\varepsilon}$ are closed and bounded in ${\mathbf{C}}$, they are compact subsets. Thus it suffices to show that $f(\gamma_{\varepsilon}(t)) \gamma_{\varepsilon}'(t)$ converges uniformly to $f(\gamma(t))\gamma'(t)$.
    • ??
• Case 3: $\Delta$ intersects both ${\mathbb{H}}^+$ and ${\mathbb{H}}^-$.
  • Break into smaller triangles, each of which falls into one of the previous two cases.

The main idea: $$\begin{align*} {\left\lvert {f'(z)} \right\rvert} &\leq {1\over 2\pi }\oint_R {{\left\lvert {f(\xi)} \right\rvert} \over {\left\lvert {\xi} \right\rvert}^2 } \,d\xi\\ &= {1\over 2\pi }\oint_R {{\left\lvert {f(\xi)} \right\rvert} } R^{-2} \,d\xi\\ &\leq {1\over 2\pi }\oint_R M R^{-2} \,d\xi\\ &= {1\over 2\pi} MR^{-2}\cdot 2\pi R \\ &= MR^{-1} \\ &\overset{R\to\infty}\longrightarrow 0 .\end{align*}$$ So $f'\equiv 0$.

Without loss of generality take $a=0$. Since $zf(z) \to A$ as $z\to 0$, $z=0$ is a simple pole of $f$ and we can write $f(z) = c_{-1}z^{-1}+ c_0 + c_1z + \cdots$. Then $$\begin{align*} \int_{\gamma_r} f(z)\,dz &= \int_{\gamma_r} \sum_{k\geq -1} c_k z^k \,dz\\ &= \sum_{k\geq -1} c_k \int_{\gamma_r} z^k \,dz\\ &= c_{-1}\int_{\gamma_r}{1\over z}\,dz\\ &= c_{-1}\int_{0}^{\beta_0} {1\over re^{i t}} ire^{it} \,dt\qquad z= re^{it}, \,dz= ire^{it} \,dt\\ &= i c_{-1}\int_{0}^{\beta_0} \,dt\\ &= i c_{-1}\beta_0 .\end{align*}$$ Now use that $$\begin{align*} zf(z) = c_{-1} + c_0z + \cdots \overset{z\to 0}\longrightarrow c_{-1} = A ,\end{align*}$$ so the integral is $iA\beta_0$.