MDoubt No 000011 - GATE

GATE [EC, 2012, 2 mark]
Question : If $f(z)=C_0+C_1z^{-1}$, then $\displaystyle\oint\limits_{\text{unit circle}}\dfrac{1+f(z)}{z}dz$ is given by

(a) $2\pi C_1$
(b) $2\pi(1+C_0)$
(c) $2\pi jC_1$
(d) $2\pi j(1+C_0)$

Solution : Let $\displaystyle I=\oint\limits_{\text{unit circle}}\dfrac{1+f(z)}{z}dz$
Given $f(z)=C_0+C_1z^{-1}$
$\displaystyle I=\oint\limits_{\text{unit circle}}\dfrac{1+C_0+C_1/z}{z}dz$
$\displaystyle I=\oint\limits_{\text{unit circle}}\dfrac{z(1+C_0)+C_1}{z^2}dz$
Singularities for $I$ is $z=0$ and $z=0$ lies inside the unit circle.
$\therefore$ by Cauchy's integral theorem we have,
$I=2\pi j\times(\text{residue of }\dfrac{z(1+C_0)+C_1}{z^2}\text{ at } z=0)$
$=2\pi j\left\{\dfrac{1}{1!}\dfrac{d}{dz}z^2\dfrac{z(1+C_0)+C_1}{z^2}\right\}_{z=0}$
$=2\pi j\{1+C_0\}_{z=0}$
$\therefore$ Answer $=2\pi j(1+C_0)$
CORRECT ANSWER : D