GATE 2018 IN Q.2: solution and answer

GATE 2018 · INQ.2 · 1 mark · MCQ

Let $f_{1} (z) = z^{2}$ and $f_{2} (z) = \overset{z}{ˉ}$ be two complex variable functions. Here $\overset{z}{ˉ}$ is the complex conjugate of $z$ . Choose the correct answer.

(A)
Both $f_{1} (z)$ and $f_{2} (z)$ are analytic
(B)
Only $f_{1} (z)$ is analytic
(C)
Only $f_{2} (z)$ is analytic
(D)
Both $f_{1} (z)$ and $f_{2} (z)$ are not analytic

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Answer and solution

ANSWER

Only $f_{1} (z) = z^{2}$ is analytic. $f_{2} (z) = \overset{z}{ˉ}$ fails the Cauchy-Riemann equations at every point. Answer: (B).

Official answer: BSource: Official answer key, GATE 2018 IN (IIT Guwahati)

Write $z = x + i y$ and split each function into $f = u + i v$ . Then test the Cauchy-Riemann equations $u_{x} = v_{y}$ and $u_{y} = - v_{x}$ .
For $f_{1} (z) = z^{2} = (x + i y)^{2} = (x^{2} - y^{2}) + i (2 x y)$ , we have $u = x^{2} - y^{2}$ and $v = 2 x y$ .
$u_{x} = 2 x = v_{y}$ and $u_{y} = - 2 y = - v_{x}$ . The Cauchy-Riemann equations hold at every point and all four partial derivatives are continuous, so $f_{1}$ is analytic everywhere. It is entire, with $f_{1}^{'} (z) = 2 z$ .
For $f_{2} (z) = \overset{z}{ˉ} = x - i y$ , we have $u = x$ and $v = - y$ .
$u_{x} = 1$ but $v_{y} = - 1$ , so $u_{x} \neq = v_{y}$ at every point. The Cauchy-Riemann equations fail everywhere, so $f_{2}$ is not differentiable at any point and cannot be analytic.
Quick check: in complex form the Cauchy-Riemann equations read $\frac{\partial f}{\partial z ˉ} = 0$ . For $z^{2}$ this derivative is $0$ . For $\overset{z}{ˉ}$ it is $1 \neq = 0$ .
So only $f_{1} (z)$ is analytic, which is option (B).

KEY IDEA

A function is analytic in a region when it satisfies the Cauchy-Riemann equations $u_{x} = v_{y}$ , $u_{y} = - v_{x}$ there and its partial derivatives are continuous. Equivalently, it must not depend on $\overset{z}{ˉ}$ : $\partial f / \partial \overset{z}{ˉ} = 0$ . Every polynomial in $z$ is entire. $\overset{z}{ˉ}$ is the standard example of a function that is continuous everywhere but differentiable nowhere.

WATCH OUT FOR

Assuming $\overset{z}{ˉ}$ is analytic because it is continuous and "simple" (it is linear in $x$ and $y$ ). Continuity is not enough. For $\overset{z}{ˉ} = x - i y$ , $u_{x} = 1$ while $v_{y} = - 1$ , so the Cauchy-Riemann equations fail at every point. Some students also get a sign wrong when expanding $z^{2}$ and wrongly conclude that $u_{y} = - v_{x}$ fails for $z^{2}$ .

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GATE 2018 IN Q.2: solution and answer

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