\documentclass[twocolumn]{article} \title{Complex Analysis,\\Problem Set 3} \author{David Schwein} \date{5 October 2014} \usepackage[T1]{fontenc} \usepackage{tikz} \usepackage{fouriernc} \usepackage{amsmath,amssymb,amsthm} \newtheorem*{theorem}{Theorem} \newtheorem*{lemma}{Lemma} \newtheorem*{proposition}{Proposition} \newtheorem*{corollary}{Corollary} \newtheorem*{conjecture}{Conjecture} \newtheorem*{problem}{Problem} \newcommand{\solution}{\paragraph{Solution}} \newcommand{\R}{\mathbb R} \newcommand{\C}{\mathbb C} \newcommand{\N}{\mathbb N} \newcommand{\Z}{\mathbb Z} \newcommand{\Hom}{\operatorname{Hom}} \newcommand{\res}{\operatorname{res}} \usepackage{fancyhdr} \pagestyle{fancy} \lhead{David Schwein\\5 October 2014} \chead{} \rhead{Complex Analysis\\Problem Set 3} \lfoot{} \cfoot{\thepage} \rfoot{} \begin{document} \maketitle \begin{problem}[4, page 202] Using residue theory, show that \[ \int_{-\infty}^{+\infty}\frac{dx}{x^4 + 1} = \frac{\pi}{\sqrt2}. \] \end{problem} \solution We integrate around the semicircular region pictured. \begin{center} \begin{tikzpicture}[>=stealth] \draw[very thin] (-1.2,0) to (2.2,0); \draw[very thin] (0,-0.2) to (0,1.4); \draw[->] (1,0) arc (0:90:1); \draw (0,1) arc (90:180:1); \draw[->] (-1,0) to (0,0); \draw (0,0) to (1,0); \node at (0,1) [above left] {$\gamma_1$}; \node at (0,0) [below left] {$\gamma_2$}; \node at (1,0) [above right] {$|z|=R$}; \end{tikzpicture} \end{center} Let $\gamma=\gamma_1 + \gamma_2$, let $f(z) = 1 + z^4$, and let $\alpha = e^{\pi i/4} = (i + 1)/\sqrt2$. Since the poles are simple we have \begin{align*} \frac{1}{2\pi i}\int_\gamma \frac{dz}{f(z)} &= \res(f(z)^{-1},\alpha) + \res(f(z)^{-1},i\alpha) \\ &= \frac{1}{4\alpha^3} + \frac{1}{4(i\alpha)^3} \\ &= \frac{1}{4\alpha^3}(1 + i) = \frac{1}{2\sqrt2\,\alpha^2} = \frac{1}{2\sqrt 2\,i} \\ \implies\quad\int_\gamma \frac{dz}{f(z)} &= \frac{\pi}{\sqrt2}. \end{align*} Furthermore, since $|z^4 + 1|\geq|R^4-1|$ when $|z|=R$, we have \[ \bigg|\int_{\gamma_1}\frac{dz}{f(z)}\bigg| \leq \frac{\pi R}{|R^4-1|}, \] giving \[ \lim_{R\to\infty}\int_{\gamma_1}\frac{dz}{f(z)} = 0. \] It follows that \[ \frac{\pi}{\sqrt2}=\lim_{R\to\infty}\int_\gamma\frac{dz}{f(z)} = \lim_{R\to\infty}\int_{\gamma_2}\frac{dz}{f(z)} = \int_{-\infty}^{+\infty} \frac{dx}{x^4 + 1}. \] \begin{problem}[2, page 205] Show using residue theory that \[ \int_{-\pi}^{+\pi} \frac{d\theta}{a+b\sin\theta} =\frac{2\pi}{\sqrt{a^2-b^2}},\qquad a>b>0. \] \end{problem} \solution Let $z=e^{i\theta}$. Then, using a substitution we discussed in class, we have \begin{align*} \int_{-\pi}^{+\pi} \frac{d\theta}{a+b\sin\theta} &= \int_{|z|=1}\frac{1}{a+(b/2i)(z-z^{-1})}\cdot\frac{dz}{iz}\\ &= 2\int_{|z|=1}\frac{dz}{bz^2 + 2aiz - b}. \end{align*} Let $f(z) = bz^2 + 2aiz - b$. The roots of $f$ are \begin{align*} \omega_+ &= \frac{i}{b}(-a + \sqrt{a^2 - b^2}) \\ \omega_- &= \frac{i}{b}(-a - \sqrt{a^2 - b^2}). \end{align*} Since $|\omega_-| > a/b > 1$ and \[ |-a + \sqrt{a^2-b^2}| = \Big|\frac{b^2}{a+\sqrt{a^2-b^2}}\Big| < \frac{b^2}{a} < b, \] the root $\omega_-$ lies outside the contour and the root $\omega_+$ lies within it. Furthermore, \begin{align*} \res\big(f(z)^{-1},\omega_+\big) &= \frac{1}{f'(\omega_+)} \\ &= \frac{1}{2b\omega_+ + ai} \\ &= \frac{1}{2i}\frac{1}{\sqrt{a^2-b^2}}. \end{align*} Hence \[ \frac{\pi}{\sqrt{a^2-b^2}} =2\pi i\int_\gamma\frac{2\,dz}{f(z)} =\int_{-\pi}^{+\pi} \frac{d\theta}{a+b\sin\theta}. \] \begin{problem}[3, page 205] Show using residue theory that \[ \int_0^\pi \frac{\sin^2\theta}{a+\cos\theta}\,d\theta =\pi(a-\sqrt{a^2-1}),\qquad a>1. \] \end{problem} \solution As in the last problem, we make the substitution $z=e^{i\theta}$, giving \begin{align*} \int_0^\pi \frac{\sin^2\theta}{a+\cos\theta}\,d\theta &= \frac{1}{2}\int_0^{2\pi} \frac{\sin^2\theta}{a+\cos\theta}\,d\theta\\ &= \frac{1}{2}\int_{|z|=1} \frac{(-1/4)(z-z^{-1})^2}{a+(z+z^{-1})/2}\cdot\frac{dz}{iz}\\ &= \frac{i}{4}\int_{|z|=1}\frac{(z-z^{-1})^2}{z^2 + 2az + 1}\,dz\\ &= \frac{i}{4}\int_{|z|=1}\frac{(z^2-1)^2}{z^2(z^2+2az+1)}\,dz. \end{align*} To compute the integral, compute the residues. Let $p(z)=(z^2-1)^2$, $q(z) = z^2(z^2+2az+1)$, and $r(z) = z^2 + 2az + 1$. The roots of~$q$ are $0$, $0$, $\beta=-a-\sqrt{a^2-1}$, and $\alpha=-a+\sqrt{a^2-1}$. Clearly $|\beta|>1$, as $a>1$; in addition, \[ |\alpha| = a-\sqrt{a^2-1} = \frac{1}{a+\sqrt{a^2-1}} < 1. \] So we need only consider the poles at~$0$ and~$\alpha$. Then \begin{align*} \res\bigg(\frac{p(z)}{q(z)},0\bigg) &= \lim_{z\to0}\frac{d}{dz}\bigg(\frac{p(z)}{r(z)}\bigg) \\ &= \frac{r(0)p'(0)-p(0)r'(0)}{r(0)^2} = -2a \end{align*} and \begin{align*} \res\bigg(\frac{p(z)}{q(z)},\alpha\bigg) &= \frac{p(\alpha)}{q'(\alpha)} \\ &= \frac{(\alpha^2-1)^2}{4\alpha^3 + 6a\alpha^2 + 2\alpha} \\ &= \frac{4(a\alpha+1)^2}{2\alpha(2\alpha^2+3a\alpha+1)} \\ &= \frac{4(a^2\alpha^2+2a\alpha+1)^2}{2\alpha(\alpha^2+a\alpha)} \\ &= \frac{4(a^2-1)\alpha^2}{2\alpha^2(\alpha+a)} \\ &= \frac{2(a^2-1)}{\alpha+a} \\ &= 2\sqrt{a^2-1}. \end{align*} Hence \begin{align*} \int_{|z|=1} \frac{p(z)}{q(z)}\,dz &= 2\pi i(-2a + 2\sqrt{a^2-1}) \\ \implies \int_0^\pi \frac{\sin^2\theta}{a+\cos\theta}\,d\theta &= \pi(a - \sqrt{a^2-1}). \end{align*} \end{document}