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Using Cauchy residue Theorem evaluate
written 5.1 years ago by | modified 5.0 years ago by |
$\int_c{\frac{sin\pi z^2+cos\pi z^2}{(z-1)^2 (z-2)}dz}$ where c is the circle ${z} = 3 $
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written 5.1 years ago by | modified 5.0 years ago by |
$\int_c{\frac{sin\pi z^2+cos\pi z^2}{(z-1)^2 (z-2)}dz}$ where c is the circle ${z} = 3 $
written 5.1 years ago by |
z=1 is a ple of order 2 and z=2 is a simple pole
Residue at z = 2
$ \begin{align} &= \lim\limits_{z \to 2} (z-2) \frac{(sin\pi z^2+cos\pi z^2)}{(z-1)^2(z-2)}\\ &= \lim\limits_{z \to 2} \frac{(sin\pi z^2+cos\pi z^2)}{(z-1)^2}\\ &= \frac{(sin4\pi +cos4\pi )}{(2-1)^2}\\ &=0+1\\ \text{Residue at (z=2)=1}\\ \text{Similarly residue at (z=1)}\\ &= \lim\limits_{z \to 1} (z-2) \frac{(sin\pi z^2+cos\pi z^2)}{(z-1)^2(z-2)}\\ &= \lim\limits_{z \to 1} \frac{(sin\pi z^2+cos\pi z^2)}{(z-1)^2}\\ &= \frac{(sin\pi +cos\pi )}{(1-2)}\\ &=\frac{0-1}{-1}\\ &=1\\ \text{Residue at (z=1)=1}\\ \end{align} $