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QUESTION


Find the value of the integral of $g(z)$ around the circle
$|z-i|=2$ where

\begin{description}

\item[(a)]
$g(z)={1\over z^2+4}${\hskip2cm}

\item[(b)]
$g(z)= {1\over (z^2+4)^2}$

\end{description}



ANSWER


Here $C$ is the circle $|z-i|=2$. The singular points of $g(z)$ in
both cases are $2i$ and $-2i$, and $2i$ lies within $C$ whilst
$-2i$ lies outside $C$.  Thus

\begin{description}

\item[(a)]
$$\int_C{dz\over z^2+4}= \int_C{dz\over (z+2i)(z-2i)}=\int_C
{f(z)\over z-2i},$$ where $f(z)={1\over z+2i}.$ Thus the integral
is equal to $2\pi i(f(2i)={\pi\over 2}$

\item[(b)]
By the same method we find that the integral is equal to $\int_C
{h(z)dz\over (z-2i)^2}$ where $h(z)={1\over (z+2i)^2}$. Thus by
the Cauchy integral formula ((*) with $n=1$)the answer is $2\pi
ih^{'}(2i)={\pi\over 16}.$

\end{description}




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