Week 6 Worksheet - Solutions

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Winding Numbers

Cauchy's Theorem

2. 

   The function $f$ is holomorphic on the domain $\mathbb{C}\setminus \{1,-1\}$. We observe that $$\mathsf{wind}({\gamma }_1,-1)=1\mathsf{wind}({\gamma }_2,1)=1\mathsf{wind}({\gamma }_3,-1)=-1=\mathsf{wind}({\gamma }_3,1)$$ and that all other winding numbers are zero. Thus $$\begin{array}{rl}\mathsf{wind}({\gamma }_1,-1)+\mathsf{wind}({\gamma }_2,-1)+\mathsf{wind}({\gamma }_3,-1)& =0\\ \mathsf{wind}({\gamma }_1,1)+\mathsf{wind}({\gamma }_2,1)+\mathsf{wind}({\gamma }_3,1)& =0\end{array}$$ and Cauchy's theorem for several contours gives $$\underset{\gamma }{\int }f+\underset{{\gamma }_1}{\int }f+\underset{{\gamma }_2}{\int }f=0$$ with the answer following from rearranging the equation.

3. It is $2\pi i$. The function $f(z)=1/z$ is holomorphic on $\mathbb{C}\setminus \{0\}$ and $\mathsf{wind}(\gamma ,0)=1$. With $\eta (t)=e^{-it}$ on $[0,2\pi ]$ we have $\mathsf{wind}(\eta ,0)=-1$ so $$\underset{\gamma }{\int }f+\underset{\eta }{\int }f=0$$ by Cauchy's theorem. The second contour integral is $$\underset{0}{\overset{2\pi }{\int }}{\displaystyle \frac{1}{e^{-it}}}(-i{e}^{-it})\mathsf{d}t=-2\pi i$$ giving the answer.
4. Let $\widetilde{{\gamma }_1}$ be the reverse of ${\gamma }_1$. We have $$\begin{array}{rl}\mathsf{wind}(\widetilde{{\gamma }_1},z_1)+\mathsf{wind}({\gamma }_2,z_1)+\mathsf{wind}(\gamma ,z_1)& =0\\ \mathsf{wind}(\widetilde{{\gamma }_1},z_2)+\mathsf{wind}({\gamma }_2,z_2)+\mathsf{wind}(\gamma ,z_2)& =0\end{array}$$ so Cauchy's theorem for several contours gives $$0=\underset{\widetilde{{\gamma }_1}}{\int }f+\underset{{\gamma }_2}{\int }f+\underset{\gamma }{\int }f=-\underset{{\gamma }_1}{\int }f+\underset{{\gamma }_2}{\int }f+\underset{\gamma }{\int }f$$ and therefore $$\underset{\gamma }{\int }f=\underset{{\gamma }_1}{\int }f-\underset{{\gamma }_2}{\int }f=(3+4i)-(5+6i)=-2-2i$$
5. From the quotient rule $f$ is differentiable at every non-zero complex number. We verify from the definition that $f$ is differentiable at zero. $$\begin{array}{rl}\underset{h\to 0}{lim}{\displaystyle \frac{f(h)-f(0)}{h}}& =\underset{h\to 0}{lim}{\displaystyle \frac{\cos (h)-1}{h^2}}\\ & =\underset{h\to 0}{lim}{\displaystyle \frac{(1-{\displaystyle \frac{h^2}{2!}}+{\displaystyle \frac{h^4}{4!}}-{\displaystyle \frac{h^6}{6!}}+\cdots )-1}{h^2}}\\ & =-\underset{h\to 0}{lim}({\displaystyle \frac{1}{2!}}-{\displaystyle \frac{h^2}{4!}}+{\displaystyle \frac{h^4}{6!}}-\cdots )\\ & ={\displaystyle \frac{1}{2}}\end{array}$$ where we evaluate the limit in the last line by plugging in $h=0$. We are allowed to do this because the power series $$\sum _{n=0}^{\mathrm{\infty }}{\displaystyle \frac{(-1{)}^n}{(2n+2)!}}{z}^{2n}$$ has $R=\mathrm{\infty }$ and is therefore continuous everywhere. Since $f$ is holomorphic on all of $\mathbb{C}$ we get $$\underset{\gamma }{\int }{\displaystyle \frac{\cos (z)-1}{z}}\mathsf{d}z=\underset{\gamma }{\int }f=0$$ as $\gamma$ is closed. Finally, this gives $$\underset{\gamma }{\int }{\displaystyle \frac{\cos (z)}{z}}\mathsf{d}z=\underset{\gamma }{\int }{\displaystyle \frac{1}{z}}\mathsf{d}z=2\pi i$$