# Laurent series for integration

• Nov 25th 2011, 12:54 AM
Daniiel
Laurent series for integration
Hey,

Just going over past papers and I think I did this question alright, just wanted to double check.

The question is

http://img443.imageshack.us/img443/83/56521555.jpg

so what I did was:

Sorry there are a few bugs in my tex, c1/... is supposed to be a frac, I couldnt see why it wasnt making it a frac, the first sum is supposed to be to infinity not y, it says infinity in the code haha, and theres afew smaller ones.

$f(z)=\frac{1}{{{z}^{2}}+1}\sum\limits_{n=0}^{\inft y}{\frac{{{z}^{n-2}}}{n!}}\,\,\,and\,\,\,\frac{1}{{{z}^{2}}+1}=\fra c{1/}{{(z-i)}{(z+i)}}=\frac{i}{2(z+i)}-\frac{i}{2(z-i)}=\frac{1}{2(-iz+1)}+\frac{1}{2(-\frac{z}{i}+1)} \\ so\,\,\,\,\frac{1}{(z-i)(z+i)}=\frac{1}{2}\sum\limits_{n=0}^{\infty }{{{z}^{n}}{{i}^{n}}}+\,\frac{1}{2}\sum\limits_{n= 0}^{\infty }{{{z}^{n}}(-{{i}^{n}})=}\frac{1}{2}\sum\limits_{n=0}^{\infty }{{{z}^{n}}({{i}^{n}}}+{{(-i)}^{n}}) \\ \frac{1}{2}\sum\limits_{n=0}^{\infty }{{{z}^{n}}({{i}^{n}}}+{{(-i)}^{n}})\times \sum\limits_{n=0}^{\infty }{\frac{{{z}^{n-2}}}{n!}}\,$

$=(1+0-{{z}^{2}}+0+{{z}^{4}}+0-{{z}^{6}}+...)(\frac{1}{{{z}^{2}}}+\frac{1}{z}+\fr ac{1}{2}+\frac{z}{6}+...) \\ =\frac{1}{{{z}^{2}}}+\frac{1}{z}+\frac{1}{2}+\frac {z}{6}+....\,\,\,\,\,\,\,so\,\,residue\,\,=\,1 \\ \int_{|z|=\frac{1}{2}}{f(z)dz=2\pi i\sum{residues}}=2\pi i\times 1=2\pi i\,\,\,\,\,\sin ce\,\,\pm i\,\,not\,\,inside\,\,|z|=\frac{1}{2} \\$

With Cauchy
$\int_{|z|=\frac{1}{2}}{f(z)dz=\frac{g(z)}{{{z}^{2} }}dz}\,=g(0)\times 2\pi i\,\,where\,\,g(z)=\frac{{{e}^{z}}}{{{z}^{2}}+1} \\=2\pi i \\$

I think it might be ok, I'm just unsure about my cauchy integral, is it fine what i did? The reason im unsure is that cauchys formula doesnt have a z squared it just has a z, but I couldnt really find a way to get around it, I guess the more correct way would be to use the cauchy derivative formula. <- just ran through it quicky and got g'(0) = 1 and then the integral was = 2iPi