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 \(iA\beta_0\).