GATE 2022 ME (ME2) – Question 49
Given $z=x+iy$, $i=\sqrt{-1}$. C is a circle of radius 2 with the centre at the origin. If the contour C is traversed anticlockwise, then the value of the integral $\frac1{2\pi}\int_C\frac1{(z-i)(z+4i)}\,dz$ is _______ (round off to one decimal place).
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Correct answer: 0.19 to 0.21
Explanation
**Step 1: poles.** The integrand $\dfrac1{(z-i)(z+4i)}$ has simple poles at $z=i$ and $z=-4i$.
**Step 2: which poles lie inside $C$?** The circle has radius 2 and centre at the origin. $|i|=1<2$ is inside. $|-4i|=4>2$ is outside, so it is ignored.
**Step 3: residue at $z=i$.**
$$\text{Res}=\lim_{z\to i}(z-i)\frac1{(z-i)(z+4i)}=\frac1{i+4i}=\frac1{5i}.$$
**Step 4: the integral.** By the residue theorem, $\oint_C f\,dz=2\pi i\times\text{Res}=2\pi i\cdot\dfrac1{5i}=\dfrac{2\pi}5$.
With the prefactor $\dfrac1{2\pi}$:
$$\frac1{2\pi}\cdot\frac{2\pi}5=\frac15=\mathbf{0.2}.$$