written 5.0 years ago by | • modified 5.0 years ago |
Solution:
$\oint_c \frac{e^{z}dz}{(z^{2}+\pi^2)^2} dz $
The poles are ,
$(z^2+\pi ^2)=0$
$(z^2-\pi ^2i^2)=0$
$(z-\pi i )(z+\pi i ) = 0$
$z=\pi i $ and $ z = -\pi i$
Hence both poles are the poles of order two,
Since unit circle is at |z|=4,
Hence both the poles lies inside the circle .
Hence we have to find two residues
$\oint_c \frac{e^{z}dz}{(z^{2}+\pi^2)^2} dz $=$\oint_c \frac{e^{z}dz}{(z-\pi i)^{2}(z+\pi i)^{2}} dz $
Rsidue at $z=\pi i$
= $\frac{1}{1!} lim_{z\to \pi i} \frac{d}{dz} [(z-\pi i)^2 \frac {e^z}{(z-\pi i)^2(z+\pi i)^{2}}]$
= $ lim_{z\to \pi i} \frac{d}{dz} [ \frac {e^z}{(z+\pi i)^{2}}]$
=$lim_{z\to \pi i} \frac {(z+\pi i)^2 e^{z}- e^{z}.2 (z+\pi i)}{(z+\pi i)^{4}}$
=$lim_{z\to \pi i} (z + \pi i) {\frac {(z+\pi i) e^{z}- e^{z}.2}{(z+\pi i)^{4}}}$
=$lim_{z\to \pi i} {\frac {(z+\pi i -2) e^{z}}{(z+\pi i)^{3}}}$
=${\frac {(\pi i+\pi i -2) e^{\pi i}}{(\pi i+\pi i)^{3}}}$
=${\frac {(2\pi i -2) e^{\pi i}}{(2\pi i)^{3}}}$
=${\frac {(\pi i -1) 2e^{\pi i}}{(8\pi ^3 i^3)}}$
= ${\frac {(\pi i -1) 2e^{\pi i}}{(8\pi^3 i^3)}}$
=${\frac {(\pi i -1) 2e^{\pi i}}{(-8\pi^3 i)}}$
=$-{\frac {(\pi i -1) e^{\pi i}}{(4\pi^3 i)}}$
=$-{\frac {(\pi i +i) e^{\pi i}i}{(4\pi^3 i)}}$
But $e^{\pi i}=cos \pi +isin \pi = -1$
-(-1).[$\frac{\pi +1}{4\pi^3}]$
Residue at z=$\pi i $= $\frac{\pi + 1}{4\pi^3}$
similarly Residue at z=-$\pi i$
=$\frac{1}{1!} lim_{z\to -\pi i} \frac{d}{dz} [(z+\pi i)^2 \frac {e^z}{(z+\pi i)^2(z-\pi i)^{2}}]$
=$lim_{z\to -\pi i} \frac{d}{dz} [\frac {e^z}{(z-\pi i)^{2}}]$
=$ lim_{z\to -\pi i} {[\frac{(z-\pi i)^2 e^z - e^z 2(z-\pi i)(1)}{(z-\pi i)^{4}}}]$
=$ lim_{z\to -\pi i} (z -\pi i){[\frac { (z-\pi i) e^z -2e^z}{(z-\pi i)^{4}}}]$
=$ lim_{z\to -\pi i} {[\frac { (z-\pi i-2) e^z}{(z-\pi i)^{3}}}]$
= $lim_{z\to -\pi i} {[\frac { (z-\pi i-2) e^z}{(z-\pi i)^{3}}}]$
=${[e^{-\pi i}\frac { (-\pi i-\pi i-2)}{(-\pi i -\pi i)^{3}}}]$
= But $e^{-\pi i}= cos \pi - isin \pi=-1$
$(-1){[{ \frac{(-2\pi i -2)}{(-8\pi^3 i)}}}]$
=$-2i{[{ \frac{(\pi -i)}{(-8\pi^3 i)}}}]$
Residue at z =$-\pi i$ = =${[{\frac{(\pi -i)}{(4\pi^3 )}}}]$
$\oint_c \frac{e^{z}dz}{(z^{2}+\pi^2)^2} dz $ = $2\pi i (sum \ of \ residue)$
=$2\pi i [\frac{\pi +i}{4\pi^3}+\frac{\pi -i}{4\pi^3}$
=$2\pi i [\frac{\pi +i+\pi-i}{4\pi^3}]$
=$2\pi i [\frac{2\pi}{4\pi^3}]$
$\oint_c \frac{e^{z}dz}{(z^{2}+\pi^2)^2} dz $=$\frac{i}{\pi}$