Question

Given $f\left(z\right)=\frac{1}{z+1}-\frac{2}{z+3}$. If $C$ is a counterclockwise path in the $z-$ plane such that $|z+1|=1$, the value of $\frac{1}{2\pi j}{\oint }_{C}f\left(z\right)dz$ is

• A. $-2$
• B. $-1$
• C. $1$
• D. $2$

Answer 1

The correct option is C $1$

Method I:

$f\left(z\right)=\frac{1}{z+1}-\frac{2}{z+3}$ and $c:|z+1|=1$ (circle)
POles of $f\left(z\right)$ are $z=-1$ and $-3$ but only $z=-1$ lies inside ${}^{\prime }{c}^{\prime }$ so we will calculate residue at $z=-1$ only. i.e.$R=\underset{(z=-1)}{Resf\left(z\right)}=\underset{z\to -1}{lim}(z+1)f\left(z\right)$$={lim}_{Z\to -1}(z+1)[\frac{1}{z+1}-\frac{2}{z+3}]$
$={lim}_{Z\to -1}(1-\frac{2(z+1)}{z+3})=1$

So by C - R - T
$\frac{1}{2}{\int }_{c}f(z_{d}z=2\pi i(R)=1$
Method II:
$f\left(z\right)=\frac{1}{z+1}-\frac{1}{z+3}$
$\frac{1}{2\pi j}{\oint }_{c}f\left(z\right)dz=\frac{1}{2\pi j}{\oint }_c\frac{dz}{z+1}-\frac{1}{2\pi j}{\oint }_c\frac{2dz}{z+3}$
$c=|z+1|=1$
$z=-3$ lies outisde $c$ and $z=-1$ lies inside $c$
$\therefore \frac{1}{2\pi j}{\oint }_{c}df\left(z\right)dz=\frac{1}{2\pi j}(-1)-0$
(Using Counchy Integral Formula)
$=1$