Question

Value of intergral, $I=\underset{C}{\oint }\frac{isin\pi z^2+icos\pi z^2}{(z-1{)}^2.(z-2)}dz$ where c is the circle |z| = 3 is _________

• A. -52

Answer 1

The correct option is A -52
$Let,f\left(z\right)=\frac{\sin \pi z^2+cos\pi z^2}{(z-1{)}^2.(z-2)}$
is analytic within C except at z = 1 & z = 2
since z = 1 is a poles of order 2,

$\therefore Resf\left(1\right)=\underset{z\to 1}{lim}\frac{1}{1}\left[\frac{d\left(\right(z-1{)}^{2}f\left(z\right))}{dz}\right]$

$=\underset{z\to 1}{lim}\frac{d}{dz}\left[\frac{sin\pi z^2+cos\pi z^2}{z-2}\right]$

$=\underset{z\to 1}{lim}\left[\frac{(z-2)(2\pi zcos\pi z^2-2\pi zsin\pi z^2)-(sin\pi z^2+cos\pi z^2)}{(z-2{)}^2}\right]$

$=(-1)(-2\pi )-(-1)$
$=2\pi +1$

$Also,Resf\left(2\right)=\underset{z\to 2}{lim}\left[\right(z-2\left)f\right(z\left)\right]$
$=\underset{z\to 2}{lim}\frac{sin\pi z^2+cos\pi z^2}{(z-1{)}^2}=1$

So using Residue theorem,

$\Rightarrow I=i\times \left[2\pi i\right[Resf\left(1\right)+Resf\left(2\right)\left]\right]$
$\Rightarrow I=-2\pi (2\pi +1+1)$
$\Rightarrow I=-52.045\approx -52$