\documentclass[a4paper,12pt]{article}

\begin{document}

\parindent=0pt

QUESTION


Let $C$ denote the boundary (taken in counterclockwise sense) of
the square whose sides lie
 along the lines $x=\pm 2$ and $y=\pm 2$. Evaluate the following
integrals:

\begin{description}

\item[(a)]
 $$\int_C {{e^{-z}\over z-(\pi i/2)}dz}$$

\item[(b)]
$$\int_C{{cos z \over z(z^2+8)}dz}$$

\item[(c)]
$$\int_C {{cosh z \over z^2}dz}$$

\end{description}



ANSWER


 We shall use the Cauchy integral formulae in the following
form. $$\int_c{f(z)dz\over (z-b)^{n+1}}={2\pi if^{(n)}(b)\over
n!}\eqno(*)$$

Here $f$ is analytic within and on a closed contour $c$.

\begin{description}

\item[(a)]
$C$ is the square bounded by $x=\pm 2, y=\pm 2$ so $\pi i/2$ lies
within $C$. Thus by (*) \indent with $n=0$ we get
$\int_C{e^{-z}dz\over z-\pi/2}=2\pi ie^{-\pi i/2}=2\pi
i(-i)=2\pi.$


\item[(b)]
As $\sqrt 8$ lies outside $C$ we let $f(z)={\cos z\over z^2+8}$
 in (*). Then
$$\int_C {\cos z\over z^2+8}=2\pi if(o)=2\pi i/8=\pi i/4.$$

\item[(c)]
In (*) we now put $n=1$, $f(z)=\cosh z$ so $$\int_C {\cosh z\over
z^2}=2\pi i \sinh 0=0.$$

\end{description}




\end{document}