MDoubt No 000013 - GATE

GATE [EC, EE, IN, 2012, 1 mark]
Question : Given $\displaystyle f(z)=\dfrac{1}{z+1}-\dfrac{2}{z+3}$. If $C$ is a counter clockwise path in the $z$-plane such that $|z+1|=1$, the value of $\displaystyle\dfrac{1}{2\pi j}\oint\limits_{C}f(z)dz$ is

(a) $-2$
(b) $-1$
(c) $1$
(d) $2$

Solution : Given $f(z)=\dfrac{1}{z+1}-\dfrac{2}{z+3}$
$\implies =\dfrac{-z+1}{(z+1)(z+3)}$
$\therefore f(z)$ has poles at $-1,-3$
$-3$ lie outside the circle $|z+1|=1$
$\therefore$ by Cauchy's Integral Formula and Cauchy's Integral Theorem, we have
$\displaystyle\dfrac{1}{2\pi j}\oint\limits_{C}f(z)dz=\dfrac{1}{2\pi j}\left[\oint\limits_{C}\dfrac{1}{z+1}dz-0\right]$
$=\dfrac{2\pi j\left(\dfrac{-z+1}{z+3}\right)_{z=-1}}{2\pi j}$
$=\dfrac{-(-1)-1}{-1+3}$
$=\dfrac{2}{2}$
$=1$
CORRECT ANSWER : C