Question

The value of the integral ${\int }_C\frac{2z+5}{(z-\frac{1}{2})(z^2-4z+5)}$
over the contour $|z|=1$, taken inthe anti-clockwise direciton , would be

• A. $\frac{24\pi i}{13}$
• B. $\frac{48\pi i}{13}$
• C. $\frac{24}{13}$
• D. $\frac{12}{13}$

Answer 1

The correct option is B $\frac{48\pi i}{13}$

$I={\oint }_C\frac{2z+5}{(z-\frac{1}{2})(z^2-4z+5)}dZ$
Only Pole $Z=1/2$ lies inside the circle $|Z|=1$
By residue theorem,
$I=2\pi i\times [\text{Residue at}Z=\frac{1}{2}]$
$=2\pi i\times \frac{(2\times \frac{1}{2}+5)}{[{\left(\frac{1}{2}\right)}^2-4\times \frac{1}{2}+5]}$
$=\frac{12\pi i}{\frac{1}{4}+3}=\frac{48\pi i}{13}$