\[ \newcommand{\Arg}{\mathsf{Arg}} \newcommand{\C}{\mathbb{C}} \newcommand{\R}{\mathbb{R}} \newcommand{\N}{\mathbb{N}} \newcommand{\Z}{\mathbb{Z}} \newcommand{\Im}{\mathsf{Im}} \newcommand{\intd}{\,\mathsf{d}} \newcommand{\Re}{\mathsf{Re}} \newcommand{\ball}{\mathsf{B}} \newcommand{\wind}{\mathsf{wind}} \newcommand{\Log}{\mathsf{Log}} \]

Week 6 Worksheet - Solutions

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Winding Numbers

Cauchy's Theorem

  1. The function $f$ is holomorphic on the domain $\C \setminus \{ 1,-1 \}$. We observe that \[ \wind(\gamma_1,-1) = 1 \qquad \wind(\gamma_2,1) = 1 \qquad \wind(\gamma_3,-1) = -1 = \wind(\gamma_3,1) \] and that all other winding numbers are zero. Thus \begin{align*} \wind(\gamma_1,-1) + \wind(\gamma_2,-1) + \wind(\gamma_3,-1) &= 0 \\ \wind(\gamma_1,1) + \wind(\gamma_2,1) + \wind(\gamma_3,1) &= 0 \end{align*} and Cauchy's theorem for several contours gives \[ \int\limits_{\gamma} f + \int\limits_{\gamma_1} f + \int\limits_{\gamma_2} f = 0 \] with the answer following from rearranging the equation.

  2. It is $2 \pi i$. The function $f(z) = 1/z$ is holomorphic on $\C \setminus \{0\}$ and $\wind(\gamma,0) = 1$. With $\eta(t) = e^{-it}$ on $[0,2\pi]$ we have $\wind(\eta,0) = -1$ so \[ \int\limits_\gamma f + \int\limits_\eta f = 0 \] by Cauchy's theorem. The second contour integral is \[ \int\limits_0^{2\pi} \dfrac{1}{e^{-it}} (-ie^{-it}) \intd t = -2 \pi i \] giving the answer.
  3. Let $\tilde{\gamma_1}$ be the reverse of $\gamma_1$. We have \begin{align*} \wind(\tilde{\gamma_1},z_1) + \wind(\gamma_2,z_1) + \wind(\gamma,z_1) &= 0 \\ \wind(\tilde{\gamma_1},z_2) + \wind(\gamma_2,z_2) + \wind(\gamma,z_2) &= 0 \end{align*} so Cauchy's theorem for several contours gives \[ 0 = \int\limits_{\tilde{\gamma_1}} f + \int\limits_{\gamma_2} f + \int\limits_\gamma f = - \int\limits_{\gamma_1} f + \int\limits_{\gamma_2} f + \int\limits_\gamma f \] and therefore \[ \int\limits_\gamma f = \int\limits_{\gamma_1} f - \int\limits_{\gamma_2} f = (3+4i) - (5+6i) = -2 - 2i \]
  4. From the quotient rule $f$ is differentiable at every non-zero complex number. We verify from the definition that $f$ is differentiable at zero. \begin{align*} \lim_{h \to 0} \dfrac{f(h) - f(0)}{h} &= \lim_{h \to 0} \dfrac{\cos(h) - 1}{h^2} \\ &= \lim_{h \to 0} \dfrac{\left( 1 - \dfrac{h^2}{2!} + \dfrac{h^4}{4!} - \dfrac{h^6}{6!} + \cdots \right) - 1}{h^2} \\ &= - \lim_{h \to 0} \left( \dfrac{1}{2!} - \dfrac{h^2}{4!} +\dfrac{h^4}{6!} - \cdots \right) \\ &= \dfrac{1}{2} \end{align*} where we evaluate the limit in the last line by plugging in $h = 0$. We are allowed to do this because the power series \[ \sum_{n=0}^\infty \dfrac{(-1)^n}{(2n+2)!} z^{2n} \] has $R = \infty$ and is therefore continuous everywhere. Since $f$ is holomorphic on all of $\C$ we get \[ \int\limits_\gamma \dfrac{\cos(z) - 1}{z} \intd z = \int\limits_\gamma f = 0 \] as $\gamma$ is closed. Finally, this gives \[ \int\limits_\gamma \dfrac{\cos(z)}{z} \intd z = \int\limits_\gamma \dfrac{1}{z} \intd z = 2 \pi i \]