The value of the integral C 2z + 5(z - 12)(z2 - 4z + 5) \, dz over the contour z = 1, taken in the anti-clockwise direction, would be

GATE 2016 EE – Question 15

Engineering Mathematics · Complex Variables: Analytic functions, Cauchy integral formula, Taylor series, Laurent series, Residue theorem · 1 mark · Multiple choice

The value of the integral $\oint_C \frac{2z + 5}{\left(z - \frac{1}{2}\right)(z^2 - 4z + 5)} \, dz$ over the contour $|z| = 1$, taken in the anti-clockwise direction, would be

$\frac{24\pi i}{13}$
$\frac{48\pi i}{13}$
$\frac{24}{13}$
$\frac{12}{13}$

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Correct answer: (B) $\frac{48\pi i}{13}$

Explanation

The roots of $z^2 - 4z + 5$ are $2 \pm i$, which lie outside the unit circle. Only the pole at $z = \frac{1}{2}$ is inside. Its residue is $\frac{2(1/2) + 5}{(1/4) - 2 + 5} = \frac{6}{13/4} = \frac{24}{13}$. The integral is $2\pi i \times \frac{24}{13} = \frac{48\pi i}{13}$.