GATE 2025 IN Q.49: solution and answer

GATE 2025 · INQ.49 · 2 marks · MSQ

Choose the correct statement(s) from the following options, regarding Cauchy's theorem on complex integration $\oint_{C} f (z) d z$ where $C$ is a simple closed path in a simply connected domain $D$ .

(A)
Cauchy's theorem cannot be directly applied to conclude that $\oint_{C} f (z) d z = 0$ when $f (z) = \frac{1}{z ^{2}}$ , and $C$ is the unit circle
(B)
If $f (z)$ is analytic in $D$ , then it can be concluded that $\oint_{C} f (z) d z = 0$ for any simple closed path $C$ in $D$
(C)
The function $f (z)$ must be analytic in $D$ to conclude $\oint_{C} f (z) d z = 0$ for any simple closed path $C$ in $D$
(D)
$\oint_{C} f (z) d z \neq = 0$ when $f (z) = \frac{1}{z ^{2}}$ , since the function is not analytic at $z = 0$

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Answer and solution

ANSWER

(A) and (B) are correct. Cauchy's theorem needs f to be analytic on and inside C, and 1/z² fails this at z = 0, so the theorem cannot be applied directly (A). If f is analytic in D, the theorem gives zero (B). The integral of 1/z² around the unit circle is in fact 0 (its residue is 0), so (D) is false. For the same reason analyticity is sufficient but not necessary, so (C) is false.

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

Cauchy's integral theorem: if $f (z)$ is analytic in a simply connected domain $D$ , then $\oint_{C} f (z) d z = 0$ for every simple closed path $C$ in $D$ . Statement (B) is exactly this theorem, so (B) is true.
(A): $f (z) = \frac{1}{z ^{2}}$ is not defined at $z = 0$ , where it has a pole of order 2. A simply connected domain $D$ that contains the unit circle $∣ z ∣ = 1$ must also contain everything inside it, including $z = 0$ . So $f$ is not analytic in $D$ , the hypothesis of Cauchy's theorem fails, and the theorem cannot be used directly to conclude $\oint_{C} f (z) d z = 0$ . (A) is true.
(D), computed directly: on $∣ z ∣ = 1$ put $z = e^{i θ}$ , so $d z = i e^{i θ} d θ$ : $\oint_{C} \frac{d z}{z ^{2}} = \int_{0}^{2 π} e^{- 2 i θ} i e^{i θ} d θ = i \int_{0}^{2 π} e^{- i θ} d θ = i [\frac{e ^{- i θ}}{- i}]_{0}^{2 π} = - (e^{- 2 π i} - 1) = 0$
Check with residues: the Laurent series of $z^{- 2}$ about $0$ is just $z^{- 2}$ . It has no $z^{- 1}$ term, so $Res_{z = 0} \frac{1}{z ^{2}} = 0$ and $\oint_{C} = 2 π i \cdot 0 = 0$ . Put another way, $\frac{1}{z ^{2}}$ has the single-valued antiderivative $F (z) = - \frac{1}{z}$ for $z \neq = 0$ , so its integral around any closed path that does not pass through $0$ is $F (end) - F (start) = 0$ . In general $\oint_{∣ z ∣ = 1} z^{n} d z = 0$ for every integer $n \neq = - 1$ . The integral is zero even though $f$ is not analytic at $0$ , so (D) is false: a function that fails to be analytic at a point inside $C$ does not, for that reason alone, have a non-zero integral.
(C): Cauchy's theorem makes analyticity a sufficient condition for $\oint_{C} f (z) d z = 0$ . It is not a necessary one. Step 4 gives a counterexample: $f (z) = 1/ z^{2}$ is not analytic at $z = 0$ , yet its integral is zero around every closed path that avoids $z = 0$ , including paths that enclose it. So it is false that $f$ "must" be analytic. (Morera's theorem, a partial converse, also requires $f$ to be continuous at every point of $D$ . $1/ z^{2}$ is not even defined at $0$ , so the example does not contradict it.) (C) is false.
Correct statements: (A) and (B).

KEY IDEA

Cauchy's theorem runs one way only: if f is analytic in D, the integral is zero. The converse does not follow. A function with a singularity inside C can still have a zero integral whenever its residue there is zero. 1/z² is the standard example, because it has the single-valued antiderivative −1/z. The theorem is silent in that case, so you cannot use it, but you also cannot conclude the integral is non-zero.

WATCH OUT FOR

Reading Cauchy's theorem as "analytic if and only if the integral is zero". That reading makes (C) and (D) look true: (D) because "not analytic, therefore non-zero" and (C) because "must be analytic". The integral of 1/z² around the unit circle is 0 because the residue is 0. Some students also cite Morera's theorem to defend (C), but Morera needs f to be continuous throughout D, and 1/z² is not continuous at z = 0.

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GATE 2025 IN Q.49: solution and answer

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