Question

The value of the integral ${\int }_c\frac{cos2\pi z}{(2z-1)(z-3)}dz$ (Where $C$ is a closed curve given by $|z|=1$ is

• A. $i$
• B. $\frac{\pi i}{5}$
• C. $\frac{2\pi i}{5}$
• D. $\pi i$

Answer 1

The correct option is C $\frac{2\pi i}{5}$

${\int }_c\frac{cos2\pi z}{(2z-1)(z-3)}dz={\int }_{C}f\left(z\right)dz$
Where $\left(C\right):|Z|=1$
Only one pole $z=1/2$ lies inside the circle $C$
Res ($f\left(z\right)$ at $z=1/4$)
$={lim}_{z=\to \frac{1}{2}}\left\{(z-\frac{2}{2})f\right(z\left)\right\}$
$={lim}_{z=\to \frac{1}{2}}\left\{\frac{cos\left(2\pi z\right)}{2(z-3)}\right\}$
$=\frac{cos\left(2\pi \frac{1}{2}\right)}{2(\frac{1}{2}-3)}=\frac{1}{5}$
By Cauchy's residue's theorem,
$I=\left(2\pi i\right)\left(\frac{1}{5}\right)=\frac{2\pi i}{5}$