Solved Complex Analysis

Mar 2010
18
0
\(\displaystyle \int\frac{sin(\pi.z)}{z^{2}(z-2)}dz\)

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
Mar 2009
2,162
994
near Piacenza (Italy)
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

\(\displaystyle \chi\) \(\displaystyle \sigma\)