MDoubt No 000015 - GATE

GATE [EE, 2016, 1 mark]
Question : The value of the integral $$\oint\limits_{C}\dfrac{2z+5}{\left(z-\dfrac{1}{2}\right)(z^2-4z+5)}dz$$ over the contour $|z|=1$, taken in the anti-clockwise direction, would be

(a) $\dfrac{24\pi i}{13}$
(b) $\dfrac{48\pi i}{13}$
(c) $\dfrac{24}{13}$
(d) $\dfrac{12}{13}$

Solution : Let $f(z)=\dfrac{2z+5}{\left(z-\dfrac{1}{2}\right)(z^2-4z+5)}$
Poles of $f(z)$ are $\dfrac{1}{2},2+i$ and $2-i$
Only $z=\dfrac{1}{2}$ lies inside the circle $|z|=1$
$\displaystyle \text{Residue, }R=\lim_{z\to 1/2}\left[\left(z-\dfrac{1}{2}\right)f(z)\right]$
$\displaystyle R=\lim_{z\to 1/2}\left[\dfrac{2z+5}{z^2-4z+5}\right]$
$\displaystyle R=\dfrac{2\left(\dfrac{1}{2}\right)+5}{\left(\dfrac{1}{2}\right)^2-4\left(\dfrac{1}{2}\right)+5}$
$\displaystyle R=\dfrac{24}{13}$
$\therefore$ by Residue theorem, we get
$\displaystyle\oint\limits_{C}\dfrac{2z+5}{\left(z-\dfrac{1}{2}\right)(z^2-4z+5)}dz=2\pi iR=\dfrac{48\pi i}{13}$
CORRECT ANSWER : B