# Solved Complex Analysis

#### bobisback

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

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 $$\displaystyle \pi i$$

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

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

thanks bobisback

#### chisigma

MHF Hall of Honor
The essential point is that the integral can be written as...

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

... where $$\displaystyle sinc (z) = \frac{\sin (\pi z)}{\pi z}$$ 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 $$\displaystyle z=0$$ and $$\displaystyle z=2$$. In this case we can apply the Cauchy's integral formula...

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

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

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

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

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

Kind regards

$$\displaystyle \chi$$ $$\displaystyle \sigma$$