GATE 2023 EC Q.14: solution and answer

GATE 2023 · ECQ.14 · 1 mark · MCQ

The value of the contour integral, $\oint_{C} (\frac{z + 2}{z ^{2} + 2 z + 2}) d z$ , where the contour C is ${z : z + 1 - \frac{3}{2} j = 1}$ , taken in the counter clockwise direction, is

(A)
$- π (1 + j)$
(B)
$π (1 + j)$
(C)
$π (1 - j)$
(D)
$- π (1 - j)$

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Answer and solution

ANSWER

The integral equals $π (1 + j)$ , which is option (B).

Official answer: BSource: Official answer key, GATE 2023 EC (IIT Kanpur)

Find the poles by solving $z^{2} + 2 z + 2 = 0$ : $z = \frac{- 2 \pm 4 - 8}{2} = - 1 \pm j$ . They are distinct, so both are simple poles, and $z^{2} + 2 z + 2 = (z - (- 1 + j)) (z - (- 1 - j))$ . The numerator $z + 2$ is not zero at either of them.
Rewrite the contour as $z - (- 1 + \frac{3}{2} j) = 1$ : a circle with centre $- 1 + \frac{3}{2} j$ and radius $1$ .
Check each pole against the circle. $(- 1 + j) - (- 1 + \frac{3}{2} j) = - \frac{1}{2} j = \frac{1}{2} < 1$ , so $- 1 + j$ is inside. $(- 1 - j) - (- 1 + \frac{3}{2} j) = - \frac{5}{2} j = \frac{5}{2} > 1$ , so $- 1 - j$ is outside. Neither pole lies on $C$ .
Residue at the simple pole $z_{0} = - 1 + j$ : $Res = \frac{z + 2}{z - ( - 1 - j )}_{z = - 1 + j} = \frac{( - 1 + j ) + 2}{( - 1 + j ) + 1 + j} = \frac{1 + j}{2 j}$ .
Apply the Cauchy residue theorem (counter-clockwise, one pole inside): $\oint_{C} \frac{z + 2}{z ^{2} + 2 z + 2} d z = 2 π j \cdot \frac{1 + j}{2 j} = π (1 + j) .$
This is option (B).

KEY IDEA

Only the poles inside the contour count. The circle is centred at $- 1 + \frac{3}{2} j$ with radius 1, so it contains $- 1 + j$ (distance $\frac{1}{2}$ ) but not $- 1 - j$ (distance $\frac{5}{2}$ ). The integral is therefore $2 π j$ times the single residue $\frac{1 + j}{2 j}$ .

WATCH OUT FOR

Reading the centre as $1 - \frac{3}{2} j$ instead of $- 1 + \frac{3}{2} j$ . The condition $∣ z + 1 - \frac{3}{2} j ∣ = 1$ means $∣ z - (- 1 + \frac{3}{2} j) ∣ = 1$ , and with the wrong sign the circle encloses neither pole, which gives a wrong value of 0. A second mistake is adding the residues at both poles, which gives $2 π j$ (the full integral around a large circle) and is not one of the options.

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GATE 2023 EC Q.14: solution and answer

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