Cauchy’s Theorem

Entire and \(O\) of polynomial implies polynomial #complex/exercise/completed

Let \(f(z)\) be entire and assume that \({\left\lvert {f(z)} \right\rvert} \leq M |z|^2\) outside of some disk for some constant \(M\). Show that \(f(z)\) is a polynomial in \(z\) of degree \(\leq 2\).

  • 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 \(n\).

Uniform sequence implies uniform derivatives #complex/exercise/work

Let \(a_n(z)\) be an analytic sequence in a domain \(D\) such that \(\displaystyle \sum_{n=0}^\infty |a_n(z)|\) converges uniformly on bounded and closed sub-regions of \(D\). Show that \(\displaystyle \sum_{n=0}^\infty |a'_n(z)|\) converges uniformly on bounded and closed sub-regions of \(D\).

Tie’s Extra Questions: Spring 2014 #complex/exercise/completed

The question provides some insight into Cauchy’s theorem. Solve the problem without using the Cauchy theorem.

  • Evaluate the integral \(\displaystyle{\int_{\gamma} z^n dz}\) for all integers \(n\). Here \(\gamma\) is any circle centered at the origin with the positive (counterclockwise) orientation.

  • Same question as (a), but with \(\gamma\) any circle not containing the origin.

  • Show that if \(|a|<r<|b|\), then \(\displaystyle{\int_{\gamma}\frac{dz}{(z-a)(z-b)} dz=\frac{2\pi i}{a-b}}\). Here \(\gamma\) denotes the circle centered at the origin, of radius \(r\), with the positive orientation.

\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.

Fall 2019.3, Spring 2020 HW 2.9 (Cauchy’s Formula for Exterior Regions) #complex/qual/completed

Let \(\gamma\) be a piecewise smooth simple closed curve with interior \(\Omega_1\) and exterior \(\Omega_2\). Assume \(f'\) exists in an open set containing \(\gamma\) and \(\Omega_2\) with \(\lim_{z\to \infty} f(z) = A\). Show that \begin{align*} F(z) \coloneqq\frac{1}{2 \pi i} \int_{\gamma} \frac{f(\xi)}{\xi-z} d \xi=\left\{\begin{array}{ll} A, & \text { if } z \in \Omega_{1} \\ -f(z)+A, & \text { if } z \in \Omega_{2} \end{array}\right. .\end{align*}

NOTE (DZG): I think there is a typo in this question….probably this should equal \(f(z)\) for \(z\in \Omega_1\), which is Cauchy’s formula…

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*}

Tie’s Extra Questions: Fall 2009 (Proving Cauchy using Green’s) #complex/exercise/completed

State and prove Green’s Theorem for rectangles. Use this to prove Cauchy’s Theorem for functions that are analytic in a rectangle.

Suppose \(f\in C_{\mathbf{C}}^1(\Omega)\) and \(T\subset \Omega\) is a triangle with \(T^\circ \subset \Omega\).

  • Apply Green’s theorem to show that \(\int_T f(z) ~dz = 0\).
  • Assume that \(f'\) is continuous and prove Goursat’s theorem.

Hint: Green’s theorem states \begin{align*} \int_{T} F d x+G d y=\int_{T^\circ}\left(\frac{\partial G}{\partial x}-\frac{\partial F}{\partial y}\right) d x d y .\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*}

No polynomials converging uniformly to \(1/z\) #complex/exercise/completed

Prove that there is no sequence of polynomials that uniformly converge to \(f(z) = {1\over z}\) on \(S^1\).

  • 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 \(\int f = \int p_j = 0\).

Eventually sublinear implies constant #complex/exercise/completed

Suppose \(f: {\mathbf{C}}\to {\mathbf{C}}\) is entire and \begin{align*} {\left\lvert {f(z)} \right\rvert} \leq {\left\lvert {z} \right\rvert}^{1\over 2} \quad\text{ when } {\left\lvert {z} \right\rvert} > 10 .\end{align*}

Prove that \(f\) is constant.

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*}

The Cauchy pole function is holomorphic #complex/exercise/completed

Let \(\gamma\) be a smooth curve joining two distinct points \(a, b\in {\mathbf{C}}\).

Prove that the function \begin{align*} f(z) \coloneqq\int_\gamma {g(w) \over w-z} \,dw \end{align*} is analytic in \({\mathbf{C}}\setminus\gamma\).

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.

Schwarz reflection proof #complex/exercise/completed

Suppose that \(f: {\mathbf{C}}\to{\mathbf{C}}\) is continuous everywhere and analytic on \({\mathbf{C}}\setminus {\mathbf{R}}\) and prove that \(f\) is entire.

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.

Prove Liouville #complex/exercise/completed

Prove Liouville’s theorem: suppose \(f:{\mathbf{C}}\to{\mathbf{C}}\) is entire and bounded. Use Cauchy’s formula to prove that \(f'\equiv 0\) and hence \(f\) is constant.

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 \(f'\equiv 0\).

Tie’s Extra Questions Fall 2009 (Fractional residue formula) #complex/exercise/completed

Assume \(f\) is continuous in the region: \begin{align*} 0 < {\left\lvert {z-a} \right\rvert} \leq R,\quad 0 \leq \operatorname{Arg}(z-a) \leq \beta_0 \qquad \beta_0\in (0, 2\pi] .\end{align*}

and the following limit exists: \begin{align*} \lim_{z\to a}(z-a)f(z) = A .\end{align*} Show that \begin{align*}\lim_{r \rightarrow 0} \int_{\gamma_r} f(z) dz = i A \beta_0 \; , \; \;\end{align*} where \begin{align*} \gamma_r : = \{ z \; | \; z = a + r e^{it}, \; 0 \leq t \leq \beta_0 \}. .\end{align*}

Let \(f\) be a continuous function in the region \begin{align*} D=\{z {~\mathrel{\Big\vert}~}{\left\lvert {z} \right\rvert}>R, 0\leq \arg z\leq \theta\}\quad\text{where}\quad 1\leq \theta \leq 2\pi .\end{align*} If there exists \(k\) such that \(\displaystyle{\lim_{z\to\infty} zf(z)=k}\) for \(z\) in the region \(D\). Show that \begin{align*} \lim_{R'\to\infty} \int_{L} f(z) dz=i\theta k ,\end{align*} where \(L\) is the part of the circle \(|z|=R'\) which lies in the region \(D\).

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 \(iA\beta_0\).

Spring 2020 HW 2, 2.6.7 #complex/exercise/work

Suppose \(f: {\mathbb{D}}\to {\mathbf{C}}\) is holomorphic and let \(d \coloneqq\sup_{z, w\in {\mathbb{D}}}{\left\lvert {f(z) - f(w)} \right\rvert}\) be the diameter of the image of \(f\). Show that \(2 {\left\lvert {f'(0)} \right\rvert} \leq d\), and that equality holds iff \(f\) is linear, so \(f(z) = a_1 z + a_2\).

Hint: \begin{align*} 2f'(0) = \frac{1}{2\pi i} \int_{{\left\lvert {\xi } \right\rvert}= r} \frac{ f(\xi) - f(-\xi) }{\xi^2} ~d\xi \end{align*} whenever \(0<r<1\).

Spring 2020 HW 2, 2.6.8 #complex/exercise/work

Suppose that \(f\) is holomorphic on the strip \(S = \left\{{x+iy {~\mathrel{\Big\vert}~}x\in {\mathbf{R}},~ -1<y<1}\right\}\) with \({\left\lvert {f(z)} \right\rvert} \leq A \qty{1 + {\left\lvert {z} \right\rvert}}^\nu\) for \(\nu\) some fixed real number. Show that for all \(z\in S\), for each integer \(n\geq 0\) there exists an \(A_n \geq 0\) such that \({\left\lvert {f^{(n)}(x)} \right\rvert} \leq A_n (1 + {\left\lvert {x} \right\rvert})^\nu\) for all \(x\in {\mathbf{R}}\).

Hint: Use the Cauchy inequalities.

Spring 2020 HW 2, 2.6.9 #complex/exercise/work

Let \(\Omega \subset {\mathbf{C}}\) be open and bounded and \(\phi: \Omega \to \Omega\) holomorphic. Prove that if there exists a point \(z_0 \in \Omega\) such that \(\phi(z_0) = z_0\) and \(\phi'(z_0) = 1\), then \(\phi\) is linear.

Hint: assume \(z_0 = 0\) (explain why this can be done) and write \(\phi(z) = z + a_n z^n + O(z^{n+1})\) near \(0\). Let \(\phi_k = \phi \circ \phi \circ \cdots \circ \phi\) and prove that \(\phi_k(z) = z + ka_nz^n + O(z^{n+1})\). Apply Cauchy’s inequalities and let \(k\to \infty\) to conclude.

Spring 2020 HW 2, 6 #complex/exercise/work

Show by example that there exists a function \(f(z)\) that is holomorphic on \(\left\{{z\in {\mathbf{C}}{~\mathrel{\Big\vert}~}0 < {\left\lvert {z} \right\rvert} < 1}\right\}\) and for all \(r<1\), \begin{align*} \int_{{\left\lvert {z} \right\rvert} = r} f(z) \, dz = 0 ,\end{align*} but \(f\) is not holomorphic at \(z=0\).

Spring 2020 HW 2, 7 #complex/exercise/work

Let \(f\) be analytic on a region \(R\) and suppose \(f'(z_0) \neq 0\) for some \(z_0 \in R\). Show that if \(C\) is a circle of sufficiently small radius centered at \(z_0\), then \begin{align*} \frac{2 \pi i}{f^{\prime}\left(z_{0}\right)}=\int_{C} \frac{d z}{f(z)-f\left(z_{0}\right)} .\end{align*}

Hint: use the inverse function theorem.

Spring 2020 HW 2, 8 #complex/exercise/work

Assume two functions \(u, b: {\mathbf{R}}^2 \to {\mathbf{R}}\) have continuous partial derivatives at \((x_0 ,y_0)\). Show that \(f \coloneqq u + iv\) has derivative \(f'(z_0)\) at \(z_0 = x_0 + iy_0\) if and only if \begin{align*} \lim _{r \rightarrow 0} \frac{1}{\pi r^{2}} \int_{\left|z-z_{0}\right|=r} f(z) d z=0 .\end{align*}

Spring 2020 HW 2, 10 #complex/exercise/work

Let \(f(z)\) be bounded and analytic in \({\mathbf{C}}\). Let \(a\neq b\) be any fixed complex numbers. Show that the following limit exists: \begin{align*} \lim_{R\to \infty} \int_{{\left\lvert {z} \right\rvert} = R} {f(z) \over (z-a)(z-b)} \,dz .\end{align*}

Use this to show that \(f(z)\) must be constant.

Spring 2020 HW 2, 11 #complex/exercise/work

Suppose \(f(z)\) is entire and \begin{align*} \lim_{z\to\infty} {f(z) \over z} = 0 .\end{align*}

Show that \(f(z)\) is a constant.

Spring 2020 HW 2, 12 #complex/exercise/work

Let \(f\) be analytic in a domain \(D\) and \(\gamma\) be a closed curve in \(D\). For any \(z_0\in D\) not on \(\gamma\), show that \begin{align*} \int_{\gamma} \frac{f^{\prime}(z)}{\left(z-z_{0}\right)} d z=\int_{\gamma} \frac{f(z)}{\left(z-z_{0}\right)^{2}} d z .\end{align*} Give a generalization of this result.

Spring 2020 HW 2, 13 #complex/exercise/work

Compute \begin{align*} \int_{{\left\lvert {z} \right\rvert} = 1} \qty{z + {1\over z}}^{2n} {dz \over z} \end{align*} and use it to show that \begin{align*} \int_0^{2\pi} \cos^{2n}(\theta) \, d\theta = 2\pi \qty{1\cdot 3 \cdot 5 \cdots (2n-1) \over 2 \cdot 4 \cdot 6 \cdots (2n)} .\end{align*}

#complex/exercise/completed #complex/exercise/work #complex/qual/completed