Question

If $C$ is a cirlce $|z|=4$ and $f\left(z\right)=\frac{z^2}{(z^2-3z+2{)}^2}$ then $\oint f\left(z\right)dz$ is

• A. $1$
• B. $0$
• C. $-1$
• D. $-2$

Answer 1

The correct option is B $0$

$f\left(z\right)=\frac{z^2}{(z^2-3z+2{)}^2}$
$=\frac{z^2}{(z-1{)}^2(z-2{)}^2}$
So poles are $z=1$ and $2$
(Both are Double pole) i.e., $m=2$
$C:|Z|=4.$ Hence both the poles lies inside ${}^{\prime }{C}^{\prime }$

$=\underset{(z=1)}{\mathrm{Res}f\left(z\right)}=\frac{1}{(2-1)!}{\left(\frac{d}{\mathrm{dz}}{(z-1)}^{2}f\left(z\right)\right]}_{z=1}$
$={\left[\frac{d}{dz}\frac{z^2}{(z-2{)}^2}\right]}_{z=1}=4$
$R_2=\underset{(z=2)}{\mathrm{Res}F\left(z\right)}=\frac{1}{(2-1)!}{\left(\frac{d}{\mathrm{dz}}{(z-2)}^{2}f\left(z\right)\right]}_{Z=2}$
$=={\left[\frac{d}{dz}\frac{z^2}{(z-1{)}^2}\right]}_{z=2}=4$
So By Cauchy Residue Theorem,
$I={\int }_{C}f\left(z\right)dz=2\pi i[R_1+R_2=0]$