<html><head><meta name="color-scheme" content="light dark"></head><body><pre style="word-wrap: break-word; white-space: pre-wrap;">\documentclass[12pt]{article}
\newcommand{\ds}{\displaystyle}
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\begin{document}

{\bf Question}

\begin{itemize}
\item[i)]
Determine the radius of convergence of the series
$\ds\sum_{n=1}^\infty\frac{z^n}{n}$.

Find one point on the circle of convergence where the series
diverges, and two distinct points where the series converges.


\item[ii)]
Express $\ds\frac{z+1}{z-1}$ as a Taylor series centred at the
origin.  What is the largest region in which this series converges
to the function?

Express the same function as a Laurent series in $|z|&gt;1$.


\item[iii)]
Find all the singular points of the function

$$\frac{\left(z-\frac{\pi}{2}\right)}{(e^z-1)^3\cos z}$$

and determine their natures.

\end{itemize}



\vspace{0.25in}

{\bf Answer}

\begin{itemize}
\item[i)]
Let $\ds u_n=\frac{z^n}{n} \hspace{0.3in}
\left|\frac{u_{n+1}}{u_n}\right|=\frac{n}{n+1}|z|\to|z|$ as
$n\to\infty$  therefore $R=1$.

The series diverges at $z=1$ since $\sum\frac{1}{n}$ diverges.

At $z=-1$ the series is $\sum\frac{(-1)^n}{n}$ which is convergent
by the Leibniz test.

At $z=i$ the series is $\frac{i}{1}-\frac{1}{2}-\frac{i}{3}+
\frac{1}{4}+\frac{i}{5}-\frac{1}{6}-\cdots$

$\hspace{0.2in}=-\frac{1}{2}+\frac{1}{4}-\frac{1}{6}+\cdots+
i(1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\cdots)$

so both real and imaginary parts converge by the Leibniz test.


\item[ii)]
$\ds\frac{1}{z-1}=-(1+z+z^2+\cdots)$

so $\ds\frac{z+1}{z-1}=-(1+z)(1+z+z^2+\cdots)=
-(1+2z+2z^2+2z^3+\cdots)$

which converges for $|z|&lt;1$.

${}$

For $\ds|z|&gt;1, \hspace{0.3in}
\frac{z+1}{z-1}=\frac{z+1}{z\left(1-\frac{1}{2}\right)}$

$\ds=\left(1+\frac{1}{z}\right)
\left(1+\frac{1}{z}+\frac{1}{z^2}+\cdots\right)=
1+\frac{2}{z}+\frac{2}{z^2}+\frac{2}{z^3}+\cdots$


\item[iii)]
singularities occur at places where $e^z=1$ and $\cos z=0$

i.e. $z=2n\pi i$ and $z=(2n+1)\frac{\pi}{2}$

Now $\ds\frac{z}{e^z-1}=\frac{1}{1+\frac{z}{2!}+\cdots}\to1$ as
$z\to0$

So $\ds z^3f(z)\to\frac{-\frac{\pi}{2}}{1.1}\not=0$ as $z\to0$

Thus $f(Z)$ has a pole of order 3 at $z=0$, and by periodicity of
$e^z$ at $z=2n\pi i$.

Letting $\ds p(z)=\frac{z-(2n+1)^\frac{\pi}{2}}{\cos z}$

Use L'Hopital's rule

$\ds p(z)\to\lim_{z\to(2n+1)\frac{\pi}{2}}\frac{1}{\sin z}\not=0$
as $z\to(2n+1)\frac{\pi}{2}$

so $f(z)$ has a removable singularity at $z=\frac{\pi}{2}$, and
simple poles at $z=(2n+1)\frac{\pi}{2} \hspace{0.2in} n\in{\bf Z}
\hspace{0.2in} n\not=1$

\end{itemize}

\end{document}
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