Week 9 Worksheet - Solutions

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Poles and Residues

1. 1. $z(1-z^2)=z(1-z)(1+z)$ has simple zeroes at each of $-1,0,1$ so its reciprocal has simple poles at all three of those points. $$\begin{array}{rl}\mathsf{Res}(f,-1)& =\underset{z\to -1}{lim}(z-(-1)){\displaystyle \frac{1}{z(1-z^2)}}=\underset{z\to -1}{lim}{\displaystyle \frac{1}{z(1-z)}}=-{\displaystyle \frac{1}{2}}\\ \mathsf{Res}(f,0)& =\underset{z\to 0}{lim}(z-0){\displaystyle \frac{1}{z(1-z^2)}}=\underset{z\to 0}{lim}{\displaystyle \frac{1}{1-z^2}}=1\\ \mathsf{Res}(f,1)& =\underset{z\to 1}{lim}(z-1){\displaystyle \frac{1}{z(1-z^2)}}=\underset{z\to 1}{lim}{\displaystyle \frac{-1}{z(1+z)}}=-{\displaystyle \frac{1}{2}}\end{array}$$
   2. Since $\sin$ is non-zero at every $\pi n+\frac{\pi }{2}$ and ${\cos }^{\prime }=\sin$ we can say that the ratio has a simple pole at every $\pi n+\frac{\pi }{2}$. Now $$\mathsf{Res}(f,n\pi +\frac{\pi }{2})={\displaystyle \frac{\sin (n\pi +\frac{\pi }{2})}{-\sin (n\pi +\frac{\pi }{2})}}=-1$$ using our lemma on residues of ratios at simple poles.
   3. The numerator is non-zero at each of the simple zeroes of the denominator so the function has a simple pole at each of these zeroes. Now $$\mathsf{Res}(f,e^{ik\pi /4})=\underset{z\to e^{ik\pi /4}}{lim}(z-e^{ik\pi /4}){\displaystyle \frac{z}{1+z^4}}={\displaystyle \frac{e^{i\pi /4}}{{\displaystyle \prod _{j\ne k}(z+e^{ij\pi /4})}}}$$ but we can calculate more easily that $$\mathsf{Res}(f,e^{ik\pi /4})={\displaystyle \frac{e^{ik\pi /4}}{4e^{3ik\pi /4}}}={\displaystyle \frac{1}{4e^{ik\pi /2}}}$$ using our lemma about residues of ratios at simple poles.
   4. The denominator is $(z+i{)}^2(z-i{)}^2$ which has zeroes of order two at $i$ and $-i$. Since the numerator is non-zero at $i$ and $-i$ the ratio has poles of order two at both $i$ and $-i$. We calculate the residues using our lemma for residues at higher-order poles. $$\begin{array}{rl}\mathsf{Res}(f,i)& =\underset{z\to i}{lim}{\left({\displaystyle \frac{(z-i{)}^2(z+1{)}^2}{(z^2+1{)}^2}}\right)}^{\prime }\\ & =\underset{z\to i}{lim}{\left({\displaystyle \frac{(z+1{)}^2}{(z+i{)}^2}}\right)}^{\prime }\\ & =\underset{z\to i}{lim}{\displaystyle \frac{2(z+1)(z+i{)}^2-2(z+i)(z+1{)}^2}{(z+i{)}^4}}\\ & =\underset{z\to i}{lim}{\displaystyle \frac{2(z+1)(z+i)-2(z+1{)}^2}{(z+i{)}^3}}\\ & ={\displaystyle \frac{2(1+i)2i-2(i+1{)}^2}{(2i{)}^3}}=-{\displaystyle \frac{i}{2}}\\ \mathsf{Res}(f,-i)& =\underset{z\to -i}{lim}{\left({\displaystyle \frac{(z-(-i){)}^2(z+1{)}^2}{(z^2+1{)}^2}}\right)}^{\prime }\\ & =\underset{z\to -i}{lim}{\left({\displaystyle \frac{(z+1{)}^2}{(z-i{)}^2}}\right)}^{\prime }\\ & =\underset{z\to -i}{lim}{\displaystyle \frac{2(z+1)(z-i{)}^2-2(z-i)(z+1{)}^2}{(z-i{)}^4}}\\ & =\underset{z\to -i}{lim}{\displaystyle \frac{2(z+1)(z-i)-2(z+1{)}^2}{(z-i{)}^3}}\\ & ={\displaystyle \frac{-2(1-i)2i-2(i-1{)}^2}{(-2i{)}^3}}={\displaystyle \frac{i}{2}}\end{array}$$
2. 1. From the Taylor series $${\displaystyle \frac{\sin z}{z^2}}={\displaystyle \frac{1}{z^2}}(z-{\displaystyle \frac{z^3}{3!}}+{\displaystyle \frac{z^5}{5!}}-\cdots )={\displaystyle \frac{1}{z}}-{\displaystyle \frac{z}{3!}}+{\displaystyle \frac{z^3}{5}}-\cdots$$ has a simple pole at the origin with a residue of 1.
   2. First note that $$(\sin z{)}^2={\displaystyle \frac{1-\cos (2z)}{2}}=z^2-{\displaystyle \frac{z^4}{3}}+\cdots$$ so we can write $${\displaystyle \frac{(\sin z{)}^2}{z^4}}={\displaystyle \frac{1}{z^4}}(z^2-{\displaystyle \frac{z^4}{3}}+\cdots )={\displaystyle \frac{1}{z^2}}-{\displaystyle \frac{1}{3}}+\cdots$$ showing a pole of order two at the origin with a residue of $0$.
3. 1. On $\mathsf{Ann}(0,0,1)$ we have $${\displaystyle \frac{1}{(1-z{)}^2}}={\left({\displaystyle \frac{1}{1-z}}\right)}^{\prime }={\left(\sum _{n=0}^{\mathrm{\infty }}{z}^n\right)}^{\prime }=\sum _{n=0}^{\mathrm{\infty }}(n+1)z^n$$ so $f(z)={\displaystyle \frac{1}{z}}+2+3z+4z^2+\cdots$ thereon. Similarly, on $\mathsf{Ann}(1,0,1)$ we have $${\displaystyle \frac{1}{z}}={\displaystyle \frac{1}{1-(1-z)}}=\sum _{n=0}^{\mathrm{\infty }}(-1{)}^n(z-1{)}^n$$ so $f(z)={\displaystyle \frac{1}{(z-1{)}^2}}-{\displaystyle \frac{1}{(z-1)}}+1-(z-1)+(z-1{)}^2-\cdots$ thereon.
   2. From the above series $f$ has a simple pole at $0$ and a pole of order two at $1$. The residue at the origin is 1 and the residue at 1 is $-1$.
   3. Since the numerator is never zero and $z(1-z{)}^2$ has a simple zero at $0$; a zero of order two at $1$; we conclude that $f$ has a simple pole at $0$ and a pole of order two at $1$. The residues are $$\underset{z\to 0}{lim}{\displaystyle \frac{z}{z(1-z{)}^2}}=1$$ and $$\underset{z\to 1}{lim}{\left({\displaystyle \frac{(z-1{)}^2}{z(1-z{)}^2}}\right)}^{\prime }=\underset{z\to 0}{lim}{\left({\displaystyle \frac{1}{z}}\right)}^{\prime }=\underset{z\to 1}{lim}{\displaystyle \frac{-1}{z^2}}=-1$$ respectively, confirming our earlier answers.
4. We can write $$\begin{array}{rl}f(z)={\displaystyle \frac{z^3+3z^2+3z+1}{(z-1{)}^3}}& ={\displaystyle \frac{((z-1)+1{)}^3+3((z-1)+1{)}^2+3((z-1)+1)+1}{(z-1{)}^3}}\\ & ={\displaystyle \frac{(z-1{)}^3+6(z-1{)}^2+12(z-1)+8}{(z-1{)}^3}}\\ & ={\displaystyle \frac{8}{(z-1{)}^3}}+{\displaystyle \frac{12}{(z-1{)}^2}}+{\displaystyle \frac{6}{(z-1{)}^1}}+1\end{array}$$ so the residue at $1$ is 6.
5. The denominator has Taylor series $$z^2\sin (z)=z^2(z-{\displaystyle \frac{z^3}{3!}}+{\displaystyle \frac{z^5}{5!}}-\cdots )=z^3-{\displaystyle \frac{z^5}{3!}}+{\displaystyle \frac{z^7}{5!}}-\cdots$$ so the order of the zero at the origin is three. Since the numerator is non-zero at the origin we have a pole of order three at the origin. We therefore (taking $m=2$ in our residue lemma) must calculate $$\begin{array}{rl}& \underset{z\to 0}{lim}{\displaystyle \frac{{((z-0{)}^3{\displaystyle \frac{1}{z^2\sin (z)}})}^{(2)}}{2}}\\ =& \frac{1}{2}\underset{z\to 0}{lim}{\left({\displaystyle \frac{z}{\sin (z)}}\right)}^{(2)}\\ =& {\displaystyle \frac{1}{2}}\underset{z\to 0}{lim}{\left({\displaystyle \frac{\sin z-z\cos z}{(\sin z{)}^2}}\right)}^{\prime }\\ =& {\displaystyle \frac{1}{2}}\underset{z\to 0}{lim}{\displaystyle \frac{(\cos z-(\cos z-z\sin z))(\sin z{)}^2-2(\sin z)(\cos z)(\sin z-z\cos z)}{(\sin z{)}^4}}\\ =& {\displaystyle \frac{1}{2}}\underset{z\to 0}{lim}{\displaystyle \frac{z(\sin z{)}^3-2(\sin z{)}^2\cos z+2z(\sin z)(\cos z{)}^2}{(\sin z{)}^4}}\end{array}$$ which has the form $0/0$ if we simply plug in $z=0$. To proceed we must calculate the orders of the zeroes of the numerator and the denominator. $$\begin{array}{rl}2(\sin z{)}^2(\cos z)& =2z^2{(1-{\displaystyle \frac{z^2}{3!}}+\cdots )}^2(1-{\displaystyle \frac{z^2}{2!}}+\cdots )=2z^2-{\displaystyle \frac{5}{3}}{z}^4+\cdots \\ 2z(\sin z)(\cos z{)}^2& =2z^2(1-{\displaystyle \frac{z^2}{3!}}+\cdots ){(1-{\displaystyle \frac{z^2}{2!}}+\cdots )}^2=2z^2-{\displaystyle \frac{7}{3}}{z}^4+\cdots \\ (\sin z{)}^4& ={(z-{\displaystyle \frac{z^2}{3!}}+\cdots )}^4=z^4-\cdots \\ z(\sin z{)}^3& =z{(z-{\displaystyle \frac{z^2}{3!}}+\cdots )}^4=z^4-\cdots \end{array}$$ so we can write $$\begin{array}{rl}& {\displaystyle \frac{1}{2}}\underset{z\to 0}{lim}{\displaystyle \frac{z(\sin z{)}^3-2(\sin z{)}^2\cos z+2z(\sin z)(\cos z{)}^2}{(\sin z{)}^4}}\\ =& {\displaystyle \frac{1}{2}}\underset{z\to 0}{lim}{\displaystyle \frac{z^4-(2z^2-\frac{5}{3}{z}^4)+(2z^2-\frac{7}{3}{z}^4)}{z^4}}\\ =& {\displaystyle \frac{1}{2}}\underset{z\to 0}{lim}{\displaystyle \frac{\frac{1}{3}{z}^4}{z^4}}={\displaystyle \frac{1}{6}}\end{array}$$ finally.
6. We can write $f(z)=(z-b{)}^{m}F(z)$ and $g(z)=(z-b{)}^{n}G(z)$ where $F,G$ are holomorphic on some $\mathsf{B}(b,r)$ and both $F(b),G(b)$ are non-zero. Thus $$f(z)g(z)=(z-b{)}^{m+n}F(z)G(z)$$ with $F(z)G(z)$ holomorphic on $\mathsf{B}(b,r)$ and non-zero at $b$. The coefficient $a_0$ in the Taylor series of $F(z)G(z)$ at $b$ is therefore non-zero and from $$f(z)g(z)=(z-b{)}^{m+n}(a_0+a_1(z-b)+a_2(z-b{)}^2+\cdots )$$ we see that $f(z)g(z)$ has a zero of order $m+n$ at $b$.

Cauchy's Residue Theorem

7. 1. First factor $z^2-5z+6=(z-2)(z-3)$ to see that the integrand has simple poles at $2$ and $3$. Both are inside ${\gamma }_4$ and $\mathsf{wind}({\gamma }_4,2)=1=\mathsf{wind}({\gamma }_4,3)$ so $$\underset{{\gamma }_4}{\int }{\displaystyle \frac{1}{z^2-5z+6}}\mathsf{d}z=2\pi i\mathsf{Res}(f,2)+2\pi i\mathsf{Res}(f,3)$$ by Cauchy's residue theorem. It remains to calculate the residues $$\begin{array}{rl}\mathsf{Res}(f,2)& =\underset{z\to 2}{lim}{\displaystyle \frac{z-2}{(z-2)(z-3)}}=-1\\ \mathsf{Res}(f,3)& =\underset{z\to 3}{lim}{\displaystyle \frac{z-3}{(z-2)(z-3)}}=1\end{array}$$ giving ${\displaystyle \underset{{\gamma }_4}{\int }{\displaystyle \frac{1}{z^2-5z+6}}\mathsf{d}z=0}$.
   2. This is the same function as a) but the contour now only contains one of the poles: the simple pole at 2. Thus $$\underset{{\gamma }_4}{\int }{\displaystyle \frac{1}{z^2-5z+6}}\mathsf{d}z=2\pi i\mathsf{Res}(f,2)=-2\pi i$$
   3. Fix $a\in \mathbb{R}$. Since $\exp (az)$ is holomorphic and never zero, the integrand has simple poles at $i$ and $-i$. Both are inside ${\gamma }_2$. The residues are $$\begin{array}{rl}\mathsf{Res}(f,i)& =\underset{z\to i}{lim}{\displaystyle \frac{\exp (az)(z-i)}{(z-i)(z+i)}}={\displaystyle \frac{\exp (ai)}{2i}}={\displaystyle \frac{\cos (a)+i\sin (a)}{2i}}\\ \mathsf{Res}(f,-i)& =\underset{z\to -i}{lim}{\displaystyle \frac{\exp (az)(z+i)}{(z-i)(z+i)}}={\displaystyle \frac{\exp (-ai)}{2i}}={\displaystyle \frac{\cos (a)-i\sin (a)}{-2i}}\end{array}$$ so Cauchy's residue theorem gives $$\underset{{\gamma }_2}{\int }{\displaystyle \frac{\exp (az)}{1-z^2}}=2\pi i\mathsf{Res}(f,i)+2\pi i\mathsf{Res}(f,-i)=2\pi i\sin (a)$$ because both winding numbers are $1$.