1. ## Solved Complex Analysis

$\int\frac{sin(\pi.z)}{z^{2}(z-2)}dz$

This was the the problem i was give and it asks to integrate around the contours center a=-2, a=0, a=2 and radius 1.

i) a=-2 gives and answer of 0 because it is holomorphic everywhere

ii) a=0 gives and answer of $\pi i$

iii) a=2 gives and answer of 0 as $\int \frac{f(z)}{z-2} dz$, when $f(z)=\frac{sin(\pi.z)}{z^{2}}$, = $2\pi.i \frac{sin(\pi.2)}{2^{2}}=0$

could some please check my working for me as am a bit confused about $sin(\pi.z)$ around the contours.

thanks bobisback

2. The essential point is that the integral can be written as...

$\int_{\gamma}\frac{\sin (\pi z)}{z^{2}\cdot (z-2)}\cdot dz= \pi \int_{\gamma} \frac{sinc (z)}{z\cdot (z-2)}\cdot dz$ (1)

... where $sinc (z) = \frac{\sin (\pi z)}{\pi z}$ is an entire function [i.e. a fuction which is analytic in the whole complex plane...] , so that the function to be integrate has only single poles in $z=0$ and $z=2$. In this case we can apply the Cauchy's integral formula...

$\int_{\gamma} f(z)\cdot dz = 2\pi j \sum_{n} r_{n}$ (2)

... where the $r_{n}$ are the residues of all single poles inside $\gamma$. Now we have three possibilities...

a) if $\gamma$ is the unit circle centered in $a=-2$ there are no poles inside $\gamma$ and is $\int_{\gamma} f(z)\cdot dz =0$...

b) if $\gamma$ is the unit circle centered in $a=0$ there is a pole in $z=0$ inside $\gamma$ and its residue is $\lim_{z \rightarrow 0} \pi \frac{sinc (z)}{z-2} = - \frac{\pi}{2}$ so that is $\int_{\gamma} f(z)\cdot dz = - j \pi^{2}$...

c) if $\gamma$ is the unit circle centered in $a=2$ there is a pole in $z=2$ inside $\gamma$ and its residue is $\lim_{z \rightarrow 2} \pi \frac{sinc (z)}{z} = 0$ so that is $\int_{\gamma} f(z)\cdot dz = 0$...

Kind regards

$\chi$ $\sigma$