GATE 2025 EC Q.52: solution and answer

GATE 2025 · ECQ.52 · 2 marks · MSQ

Which of the following statements involving contour integrals (evaluated counter-clockwise) on the unit circle $C$ in the complex plane is/are TRUE?

(A)
$\oint_{C} e^{z} d z = 0$
(B)
$\oint_{C} z^{n} d z = 0$ , where $n$ is an even integer
(C)
$\oint_{C} cos z d z \neq = 0$
(D)
$\oint_{C} sec z d z \neq = 0$

See the question in the paper

Tap to open full size

Answer and solution

ANSWER

(A) and (B) are true. $e^{z}$ is entire, so its integral round the unit circle is $0$ , and $\oint_{C} z^{n} d z$ is nonzero only for $n = - 1$ , which is odd. $cos z$ is entire and every pole of $sec z$ lies at $∣ z ∣ \geq π /2 > 1$ , so both of those integrals are $0$ and statements (C) and (D) are false.

Official answer: A, BSource: Official answer key, GATE 2025 EC (IIT Roorkee)

Tool: Cauchy's integral theorem. If $f (z)$ is analytic on and inside the closed contour $C$ (here $∣ z ∣ = 1$ ), then $\oint_{C} f (z) d z = 0$ . So for each option we look for singularities inside $∣ z ∣ < 1$ .
(A) $e^{z}$ is entire (it has no singularity anywhere), so by Cauchy's theorem $\oint_{C} e^{z} d z = 0$ . TRUE.
(B) Cauchy's theorem does not settle negative $n$ , because then $z^{n}$ has a pole at $z = 0$ . So evaluate directly. Parametrise $z = e^{j θ}$ , $d z = j e^{j θ} d θ$ , $0 \leq θ \leq 2 π$ : $\oint_{C} z^{n} d z = j \int_{0}^{2 π} e^{j (n + 1) θ} d θ = {2 π j, 0, n = - 1 n \neq = - 1$ For $n \neq = - 1$ the antiderivative $\frac{e ^{j (n + 1) θ}}{j ( n + 1 )}$ takes the same value at $θ = 0$ and $θ = 2 π$ , so the integral is $0$ .
The only nonzero case is $n = - 1$ , which is odd. So for every even $n$ the integral is $0$ . That includes $n = 0$ and negative even values such as $n = - 2$ , where $z^{- 2}$ has a double pole at $0$ with residue $0$ . TRUE.
(C) $cos z$ is entire, so $\oint_{C} cos z d z = 0$ . The statement " $\neq = 0$ " is FALSE.
(D) $sec z = \frac{1}{cos z}$ is singular only where $cos z = 0$ . Write $z = x + j y$ : $cos z = cos x cosh y - j sin x sinh y$ Setting the real part to zero gives $cos x = 0$ (since $cosh y \geq 1$ ). Then $sin x = \pm 1$ , and setting the imaginary part to zero gives $sinh y = 0$ , so $y = 0$ . So every zero of $cos z$ is real: $z = \pm \frac{π}{2}, \pm \frac{3 π}{2}, \dots$
The poles nearest the origin have $∣ z ∣ = \frac{π}{2} \approx 1.571 > 1$ , so every pole lies outside $C$ . That makes $sec z$ analytic on and inside $∣ z ∣ = 1$ , and Cauchy's theorem gives $\oint_{C} sec z d z = 0$ . The statement " $\neq = 0$ " is FALSE.
Correct options: (A) and (B).

KEY IDEA

By Cauchy's theorem, a contour integral is zero whenever the integrand is analytic on and inside the contour. So first find where the singularities are. For powers of $z$ , use the standard result $\oint_{∣ z ∣ = 1} z^{n} d z = 2 π j$ if $n = - 1$ and $0$ for every other integer $n$ . A pole inside the contour does not by itself make the integral nonzero; only the residue counts.

WATCH OUT FOR

Two traps. First, treating (B) as false because a negative even power such as $z^{- 2}$ has a pole inside $C$ : that pole's residue is $0$ , so the integral is still $0$ . Second, marking (D) true because $sec z$ has poles at all: they sit at $\pm π /2, \pm 3 π /2, \dots$ (all real, since $cos z$ has no non-real zeros), which is outside the unit circle. Also watch the $\neq =$ sign in (C) and (D): both integrals are $0$ , so those statements are false.

From the GATE 2025 EC paper. Every question of this paper · Download the full paper · All 24 Complex analysis questions, with notes

GATE 2025 EC Q.52: solution and answer

Answer and solution

More Complex analysis questions