MDoubt No 000009 - GATE

GATE [CE, 2009, 2 marks]
Question : The value of the integral $\displaystyle\int\limits_{C}\dfrac{\cos (2\pi z)}{(2z-1)(z-3)}dz$ (where $C$ is a closed curve given by $|z|=1$ is

(a) $-\pi i$
(b) $\displaystyle\dfrac{\pi i}{5}$
(c) $\displaystyle\dfrac{2\pi i}{5}$
(d) $\pi i$

Solution : $\displaystyle\int\limits_{C}\dfrac{\cos (2\pi z)}{(2z-1)(z-3)}dz$
putting $(2z-1)(z-3)=0$
so the singularities are $z=\dfrac{1}{2}$ and $z=3$
but only $z=\dfrac{1}{2}$ lies inside the circle $|z|=1$ (the closed curve)
hence by Cauchy's integral theorem
$\displaystyle\dfrac{1}{2}\int\limits_{C}\dfrac{\left[\dfrac{\cos (2\pi z)}{(z-3)}\right]}{\left(z-\dfrac{1}{2}\right)}dz=\dfrac{1}{2}2\pi if(\dfrac{1}{2})$
where $f(z)=\dfrac{\cos (2\pi z)}{(z-3)}$
$\therefore\displaystyle\int\limits_{C}\dfrac{\cos (2\pi z)}{(2z-1)(z-3)}dz=\dfrac{2\pi i}{5}$
CORRECT ANSWER : C