GATE 2023 IN Q.39: solution and answer

GATE 2023 · INQ.39 · 2 marks · MCQ

Let $f (z) = j \frac{1 - z}{1 + z}$ , where $z$ denotes a complex number and $j$ denotes $- 1$ . The inverse function $f^{- 1} (z)$ maps the real axis to the ________.

(A)
unit circle with centre at the origin
(B)
unit circle with centre not at the origin
(C)
imaginary axis
(D)
real axis

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Answer and solution

ANSWER

$f^{- 1} (w) = \frac{j - w}{j + w}$ sends every real $u$ to a point with $∣ z ∣ = 1$ , so it maps the real axis onto the unit circle with centre at the origin. Answer: (A).

Official answer: ASource: Official answer key, GATE 2023 IN (IIT Kanpur)

Write $w = f (z) = j \frac{1 - z}{1 + z}$ . This is a bilinear (Mobius) map. Solve it for $z$ to get the inverse: $w (1 + z) = j (1 - z) \Rightarrow w + w z = j - j z \Rightarrow z (w + j) = j - w$ .
So the inverse map is $f^{- 1} (w) = \frac{j - w}{j + w} .$
Substitute a point of the real axis, $w = u$ with $u$ real: $z = \frac{j - u}{j + u}$ .
Take the modulus. The numerator is $- u + j$ and the denominator is $u + j$ , and both have modulus $u^{2} + 1$ : $∣ z ∣ = \frac{∣ j - u ∣}{∣ j + u ∣} = \frac{u ^{2} + 1}{u ^{2} + 1} = 1 for every real u .$
So every image point lies on $∣ z ∣ = 1$ . Check a few points: $u = 0 \to z = \frac{j}{j} = 1$ ; $u = 1 \to z = \frac{- 1 + j}{1 + j} = \frac{( - 1 + j ) ( 1 - j )}{( 1 + j ) ( 1 - j )} = \frac{2 j}{2} = j$ ; $u \to \pm \infty \to z \to - 1$ . All three lie on the unit circle. A Mobius map is one-to-one, so as $u$ runs over the whole real axis (with the point at infinity) the image covers the whole circle.
A bilinear map sends a line to either a line or a circle. The inverse map's only pole is at $w = - j$ , which is not on the real axis, so the image is bounded: it is a circle, not a line. That circle is $∣ z ∣ = 1$ , whose centre is the origin. Answer: (A).

KEY IDEA

Invert the bilinear map and substitute a real value. For real $u$ , the numbers $j - u$ and $j + u$ have the same modulus $u^{2} + 1$ , so their ratio always has modulus 1. Put the other way round: $f$ is a Cayley-type transform that takes the unit circle onto the real axis, so $f^{- 1}$ takes the real axis back onto the unit circle.

WATCH OUT FOR

Using the forward map $f$ instead of $f^{- 1}$ . For real $x$ , $f (x) = j \frac{1 - x}{1 + x}$ is purely imaginary, so $f$ maps the real axis to the imaginary axis, which leads to the wrong option (C). A second slip is getting the image circle right but not checking its centre. The points $1$ , $j$ and $- 1$ are all at distance 1 from the origin, which fixes the centre at $0$ and rules out (B).

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GATE 2023 IN Q.39: solution and answer

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