GATE 2026 EE Q.62: solution and answer

GATE 2026 · EEQ.62 · 2 marks · NAT

The magnitude of the contour integral

\oint_{C} (\frac{( z + 1 ) ^{2}}{( z - i ) ( z - 2 )}) d z

over the contour $C : ∣ z - 2 - i ∣ = 3/2$ is ______

(Round off to two decimal places)

Note: $z$ is a complex variable and $i = - 1$

See the question in the paper

Tap to open full size

Answer and solution

ANSWER

The magnitude of the contour integral is $\frac{18 π}{5} \approx 25.29$ .

Official answer: 25.00 to 25.60Source: Official answer key, GATE 2026 EE (IIT Guwahati)

The contour $C : ∣ z - 2 - i ∣ = \frac{3}{2}$ is the circle with centre $z_{0} = 2 + i$ and radius $\frac{3}{2}$ . The integrand $f (z) = \frac{( z + 1 ) ^{2}}{( z - i ) ( z - 2 )}$ has simple poles at $z = i$ and $z = 2$ . The numerator does not vanish at either point, so neither pole cancels.
Check where each pole lies. For $z = i$ : $∣ i - 2 - i ∣ = ∣ - 2∣ = 2 > \frac{3}{2}$ , so it is outside $C$ . For $z = 2$ : $∣2 - 2 - i ∣ = ∣ - i ∣ = 1 < \frac{3}{2}$ , so it is inside $C$ .
Residue at the simple pole $z = 2$ : $Res_{z = 2} f = z \to 2 lim (z - 2) f (z) = \frac{( 2 + 1 ) ^{2}}{2 - i} = \frac{9}{2 - i} = \frac{9 ( 2 + i )}{( 2 - i ) ( 2 + i )} = \frac{9 ( 2 + i )}{5}$
By the Cauchy residue theorem, only the enclosed pole counts: $\oint_{C} f (z) d z = 2 π i \cdot \frac{9 ( 2 + i )}{5} = \frac{18 π}{5} (- 1 + 2 i)$
Take the magnitude, using $∣ i ∣ = 1$ and $∣2 + i ∣ = 5$ : $\oint_{C} f (z) d z = 2 π \cdot \frac{9 5}{5} = \frac{18 π}{5} = \frac{56.5487}{2.23607} \approx 25.289$
Rounded to two decimal places, the magnitude is $25.29$ .

KEY IDEA

Find where each pole sits relative to the circle $∣ z - (2 + i) ∣ = 3/2$ by working out its distance from the centre $2 + i$ . Only $z = 2$ (distance 1) is inside, so the integral is $2 π i$ times that one residue, $\frac{9}{2 - i}$ . Its magnitude is $2 π \cdot \frac{9}{∣2 - i ∣} = \frac{18 π}{5}$ .

WATCH OUT FOR

Including the pole at $z = i$ as well. It is at distance 2 from the centre $2 + i$ , so it lies outside the radius- $3/2$ circle. Counting both residues gives $2 π i [Res_{z = i} + Res_{z = 2}]$ , which is a different number. A second slip is to treat $∣ z - 2 - i ∣$ as a circle centred at $2 - i$ or $- 2 - i$ instead of $2 + i$ . A third is to report $2 π \cdot 9/5$ after dropping the modulus of $(2 + i)$ from the residue.

From the GATE 2026 EE paper. Every question of this paper · Download the full paper · All 24 Complex analysis questions, with notes

GATE 2026 EE Q.62: solution and answer

Answer and solution

More Complex analysis questions