GATE Papers >> EEE >> 2016 >> Question No 15

Question No. 15 EEE | GATE 2016

The value of the integral

$\oint_c\frac{2z+5}{\left(z-{\displaystyle\frac12}\right)\left(z^2-4z+5\right)}dz$

over the contour z=1, taken in the anti-clockwise direction, would be


Answer : (B) 48πi13


Solution of Question No 15 of GATE 2016 EEE Paper

$ \mathrm I=\oint_\mathrm C\frac{2\mathrm z+5}{\left\{{\displaystyle\mathrm z-\frac12}\right\}{\displaystyle\left\{\mathrm z^2-4\mathrm z+5\right\}}}\mathrm{dz} $

Pole Z = $ \frac12 $ lies inside circle |z| = 1

I = 2$\mathrm\pi$ix {Residue of z at z = $\frac12$}

$ \mathrm I=2\mathrm{πi}\frac{\left\{2\times{\displaystyle\frac12}+5\right\}}{\left\{{\displaystyle\left({\displaystyle\frac12}\right)^2}{\displaystyle-}{\displaystyle4}{\displaystyle\times}{\displaystyle\frac12}{\displaystyle+}{\displaystyle5}\right\}}=\frac{12\mathrm{πi}}{\displaystyle\frac14+3}=\frac{48\mathrm{πi}}{13} $

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