ans8.tex

\begin{Solution}{8.1}
	For zeros
	\[\frac{z-2}{z^2}\sin{1 \over z-1} = 0\]
	\[z-2 = 0 \text{ and } \sin{1 \over z-1} = 0\]
	Therefore $z=2$ and $\frac{1}{z-1}=n\pi$, i.e.
	\[z= 2 \text{ and } z=\frac{1}{n\pi}+1\]
	It is obvious that $z=2$ is simple zero. For $z=\frac{1}{n\pi}+1$, limit point of zero when $n\tends \infty$ is 1, i.e., $z=1$ is isolated essential singularity. Except these $z^2=0$ provides pole at $z=0$ of order 2.
	
\end{Solution}
\begin{Solution}{8.2.b}
	Poles: $\sin z - \cos z=0 \Rightarrow z=\frac{\pi}{4}$
	
\end{Solution}
\begin{Solution}{8.2.d}
	Here $\ds f(z)=\frac{\cot \pi z}{(z-a)^2}=\frac{\cos \pi z}{\sin \pi z(z-a)^2}$, The poles are given by putting denominator equal to zero. i.e.,
	\[\sin \pi z(z-a)^2 = 0\]
	\[(z-a)^2 = 0 \Rightarrow z=a \text{ Double pole}\]
	and
	\[\sin \pi z = 0 \Rightarrow \pi z = n\pi \Rightarrow z=n\]
	Limit point of pole $z=n$, when $n\tends \infty$ is $\infty$. Thus $z=\infty$ is non-isolated essential singularity.
	
\end{Solution}
\begin{Solution}{8.2.j}
	Non-isolated essential singularity.
	
\end{Solution}