Week 9 Worksheet - Solutions

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Laurent Series

1. 1. ${\displaystyle {\displaystyle \frac{1}{z-3}}={\displaystyle \frac{1}{z}}{\displaystyle \frac{1}{1-\frac{3}{z}}}={\displaystyle \frac{1}{z}}\sum _{n=0}^{\mathrm{\infty }}{\left(\frac{3}{z}\right)}^n={\displaystyle \frac{1}{z}}+{\displaystyle \frac{3}{z^2}}+{\displaystyle \frac{9}{z^3}}+\cdots }$ whenever $\left|{\displaystyle \frac{3}{z}}\right|<1$ i.e. $|z|>3$.
   2. ${\displaystyle {\displaystyle \frac{1}{z(1-z)}}={\displaystyle \frac{1}{z}}\sum _{n=0}^{\mathrm{\infty }}{z}^n={\displaystyle \frac{1}{z}}+1+z+z^2+\cdots }$ whenever $|z|<1$.
   3. ${\displaystyle z^3\exp (1/z)=z^3\sum _{n=0}^{\mathrm{\infty }}{\displaystyle \frac{1}{n!z^n}}=z^3+z^2+{\displaystyle \frac{z}{2!}}+{\displaystyle \frac{1}{3!}}+{\displaystyle \frac{1}{4!z}}+\cdots }$ on $\mathbb{C}\setminus \{0\}$. \item ${\displaystyle \cos (1/z)=\sum _{n=0}^{\mathrm{\infty }}{\displaystyle \frac{(-1{)}^n}{(2n)!z^{2n}}}=1-{\displaystyle \frac{1}{2!z^2}}+{\displaystyle \frac{1}{4!z^4}}-{\displaystyle \frac{1}{6!z^6}}+\cdots }$ on $\mathbb{C}\setminus \{0\}$.

2. First note that $${\displaystyle \frac{1}{z+1}}=\{\begin{array}{ll}{\displaystyle {\displaystyle \frac{1}{1-(-z)}}=\sum _{n=0}^{\mathrm{\infty }}(-z{)}^n}& |z|<1\\ {\displaystyle \frac{1}{z}}{\displaystyle \frac{1}{1-(-\frac{1}{z})}}={\displaystyle \frac{1}{z}}{\displaystyle \sum _{n=0}^{\mathrm{\infty }}{(-{\displaystyle \frac{1}{z}})}^n}& |z|>1\end{array}$$ and that $${\displaystyle \frac{1}{z-3}}=\{\begin{array}{ll}-{\displaystyle \frac{1}{3}}{\displaystyle \frac{1}{1-\frac{z}{3}}}={\displaystyle -{\displaystyle \frac{1}{3}}\sum _{n=0}^{\mathrm{\infty }}{\left({\displaystyle \frac{z}{3}}\right)}^n}& |z|<3\\ {\displaystyle \frac{1}{z}}{\displaystyle \frac{1}{1-\frac{3}{z}}}={\displaystyle \frac{1}{z}}{\displaystyle \sum _{n=0}^{\mathrm{\infty }}{\left({\displaystyle \frac{3}{z}}\right)}^n}& |z|>3\end{array}$$ using geometric series.

   ${\displaystyle \sum _{n=0}^{\mathrm{\infty }}(-z{)}^n-{\displaystyle \frac{1}{3}}\sum _{n=0}^{\mathrm{\infty }}{\left({\displaystyle \frac{z}{3}}\right)}^n}$

   ${\displaystyle {\displaystyle \frac{1}{z}}\sum _{n=0}^{\mathrm{\infty }}{(-{\displaystyle \frac{1}{z}})}^n-{\displaystyle \frac{1}{3}}\sum _{n=0}^{\mathrm{\infty }}{\left({\displaystyle \frac{z}{3}}\right)}^n}$

   ${\displaystyle {\displaystyle \frac{1}{z}}\sum _{n=0}^{\mathrm{\infty }}{(-{\displaystyle \frac{1}{z}})}^n+{\displaystyle \frac{1}{z}}{\displaystyle \sum _{n=0}^{\mathrm{\infty }}{\left({\displaystyle \frac{3}{z}}\right)}^n}}$

3. 1. ${\displaystyle {\displaystyle \frac{1}{z^2(z-1)}}=-{\displaystyle \frac{1}{z^2}}{\displaystyle \frac{1}{1-z}}=-{\displaystyle \frac{1}{z^2}}\sum _{n=0}^{\mathrm{\infty }}{z}^n={\displaystyle \frac{1}{z^2}}+{\displaystyle \frac{1}{z}}+1+z+z^2+z^3+\cdots }$ on $\mathsf{Ann}(0,0,1)$
   2. First note that $${\displaystyle \frac{1}{z}}={\displaystyle \frac{1}{1-(1-z)}}=\sum _{n=0}^{\mathrm{\infty }}(1-z{)}^n$$ whenever $|z-1|<1$. Differentiating gives $$-{\displaystyle \frac{1}{z^2}}=\sum _{n=0}^{\mathrm{\infty }}n(1-z{)}^{n-1}=\sum _{n=0}^{\mathrm{\infty }}(n+1)(1-z{)}^n$$ whenever $0<|z-1|<1$. Then calculate that $$\begin{array}{rl}{\displaystyle \frac{1}{z^2(1-z)}}={\displaystyle \frac{1}{z-1}}(-{\displaystyle \frac{1}{z^2}})& ={\displaystyle \frac{1}{z-1}}\sum _{n=0}^{\mathrm{\infty }}(n+1)(z-1{)}^n\\ & ={\displaystyle \frac{1}{z-1}}+2+3(z-1)+4(z-1{)}^2+\cdots \end{array}$$ for $0<|z-1|<1$.
4. 1. Just ${\displaystyle \frac{1}{(z-1{)}^2}}$
   2. ${\displaystyle \frac{1}{z-1}}=-{\displaystyle \sum _{n=0}^{\mathrm{\infty }}{z}^n}$ for $|z|<1$ and so $$\frac{1}{(z-1{)}^2}=-{\left({\displaystyle \frac{1}{z-1}}\right)}^{\prime }=-\sum _{n=0}^{\mathrm{\infty }}n{z}^{n-1}=-\sum _{n=0}^{\mathrm{\infty }}(n+1)z^n$$ on $\mathsf{Ann}(0,0,1)$. (Note that $f$ is holomorphic on $\mathsf{B}(0,1)$.)
   3. From part b) we have $$-\sum _{n=0}^{\mathrm{\infty }}{\displaystyle \frac{n+1}{z^n}}={\displaystyle \frac{1}{(\frac{1}{z}-1{)}^2}}=z^2{\displaystyle \frac{1}{(1-z{)}^2}}$$ whenever $|z|>1$. Thus $${\displaystyle \frac{1}{(1-z{)}^2}}={\displaystyle \frac{1}{z^2}}\sum _{n=0}^{\mathrm{\infty }}{\displaystyle \frac{n+1}{z^n}}$$ on $\mathsf{Ann}(0,1,\mathrm{\infty })$.

Isolated Singularities

5. 1. $z^2+1=(z+i)(z-i)$ has simple zeroes at $i$ and $-i$ so $f$ has simple poles at $i$ and $-i$.
   2. $z^4+16=(z-2e^{i\pi /4})(z-2e^{3i\pi /4})(z-2e^{5i\pi /4})(z-2e^{7\pi i/4})$ has simple zeroes at $2e^{i\pi /4},2e^{3i\pi /4},2e^{5i\pi /4},2e^{7i\pi /4}$ so $f$ has simple zeroes at each of those points.
   3. $z^4+2z^2+1=(z^2+1{)}^2=(z+i{)}^2(z-i{)}^2$ has zeroes of order two at $i$ and $-i$ so $f$ has poles of order two at $i$ and $-i$.
   4. $z^2+z-1=(z-{\displaystyle \frac{-1-\sqrt{5}}{2}})(z-{\displaystyle \frac{-1+\sqrt{5}}{2}})$ has simple zeroes so $f$ has simple poles at ${\displaystyle \frac{-1-\sqrt{5}}{2}}$ and ${\displaystyle \frac{-1+\sqrt{5}}{2}}$.
6. 1. Since $$\sin (1/z)={\displaystyle \sum _{n=0}^{\mathrm{\infty }}{\displaystyle \frac{(-1{)}^n}{(2n+1)!}}{\displaystyle \frac{1}{z^n}}}$$ has infinitely many non-zero coefficients the singularity is essential.
   2. The function $g(z)=(\sin (z){)}^2$ is holomorphic on all of $\mathbb{C}$ so is represented by a power series $$(\sin (z){)}^2=\sum _{n=0}^{\mathrm{\infty }}{a}_n{z}^n$$ with infinite radius of convergence. However $(\sin z{)}^2$ has a zero of order two at the origin: the coefficients $a_0$ and $a_1$ are both zero whereas $a_2$ is non-zero. Indeed, calculating derivatives $$((\sin z{)}^2{)}^{\prime }=2(\sin z)(\cos z)((\sin z{)}^2{)}^{″}=2(\cos z{)}^2-2(\sin z{)}^2$$ we see that the first derivative is zero at the origin while the second is non-zero at the origin. Thus $$\begin{array}{rl}{\displaystyle \frac{1}{z^3}}(\sin z{)}^2& ={\displaystyle \frac{1}{z^3}}(a_2{z}^2+a_3{z}^3+\cdots )\\ & ={\displaystyle \frac{a_2}{z}}+a_3+a_{4}z+a_5{z}^2+\cdots \end{array}$$ with $a_2$ non-zero so we have a simple pole at the origin.
   3. From our power series for cosine $$\begin{array}{rl}{\displaystyle \frac{1-\cos (z)}{z^2}}& ={\displaystyle \frac{1}{z^2}}(1-(1-{\displaystyle \frac{z^2}{2!}}+{\displaystyle \frac{z^4}{4!}}-{\displaystyle \frac{z^6}{6!}}+\cdots ))\\ & =\frac{1}{z^2}({\displaystyle \frac{z^2}{2!}}-{\displaystyle \frac{z^4}{4!}}+{\displaystyle \frac{z^6}{6!}}-\cdots )\\ & ={\displaystyle \frac{1}{2!}}-{\displaystyle \frac{z^2}{4!}}+{\displaystyle \frac{z^4}{6!}}-\cdots \end{array}$$ so the singularity at the origin is removable.
7. Fix $M>0$ such that $|f(z)|\le M$ for all $z\in D$. Fix $R>0$ such that $\mathsf{B}(b,R)\subset D$. Since $f$ is holomorphic on $\mathsf{Ann}(b,0,R)$ it has a Laurent series thereon. From Laurent's theorem $$a_{-n}={\displaystyle \frac{1}{2\pi i}}\underset{\gamma }{\int }{\displaystyle \frac{f(z)}{(z-b{)}^{-n+1}}}\mathsf{d}z$$ where $\gamma (t)=b+r{e}^{it}$ for every $0<r<R$ and every $n\in \mathbb{N}$. Thus $$\begin{array}{rl}|a_{-n}|& ={\displaystyle \frac{1}{2\pi }}\left|\underset{\gamma }{\int }{\displaystyle \frac{f(z)}{(z-b{)}^{-n+1}}}\mathsf{d}z\right|\\ & \le {\displaystyle \frac{1}{2\pi }}2\pi r{\displaystyle \frac{B}{r^{-n+1}}}=B{r}^n\end{array}$$ from our estimation lemma. Since $r>0$ was arbitrary we conclude that $a_{-n}=0$ for all $n\in \mathbb{N}$. Thus the singularity is removable.