GATE 2024 IN Q.11: solution and answer

GATE 2024 · INQ.11 · 1 mark · MCQ

Let $z = x + i y$ be a complex variable and $\overset{z}{ˉ}$ be its complex conjugate. The equation $\overset{z}{ˉ}^{2} + z^{2} = 2$ represents a

(A)
parabola
(B)
hyperbola
(C)
ellipse
(D)
circle

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Answer and solution

ANSWER

$\overset{z}{ˉ}^{2} + z^{2} = 2$ reduces to $x^{2} - y^{2} = 1$ , a rectangular hyperbola, so the answer is (B).

Official answer: BSource: Official answer key, GATE 2024 IN (IISc Bengaluru)

Write $z = x + i y$ and $\overset{z}{ˉ} = x - i y$ , then expand both squares: $z^{2} = x^{2} - y^{2} + 2 i x y$ and $\overset{z}{ˉ}^{2} = x^{2} - y^{2} - 2 i x y$ .
Add them. The imaginary parts $\pm 2 i x y$ cancel, which is expected because $\overset{z}{ˉ}^{2}$ is the conjugate of $z^{2}$ , so $z^{2} + \overset{z}{ˉ}^{2} = 2 Re (z^{2})$ : $z^{2} + \overset{z}{ˉ}^{2} = 2 (x^{2} - y^{2})$
Set this equal to 2: $2 (x^{2} - y^{2}) = 2 \Rightarrow x^{2} - y^{2} = 1$ .
Compare with the standard form $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$ . Here $a = b = 1$ , so the curve is a rectangular hyperbola with vertices at $(\pm 1, 0)$ and asymptotes $y = \pm x$ .
So the equation represents a hyperbola, which is option (B).

KEY IDEA

Substitute $z = x + i y$ and turn the complex equation into a real equation in $x$ and $y$ . A sum of conjugates is real: $z^{2} + \overset{z}{ˉ}^{2} = 2 Re (z^{2}) = 2 (x^{2} - y^{2})$ . Then read off the conic from its Cartesian form: $x^{2} - y^{2} = 1$ is a hyperbola.

WATCH OUT FOR

Treating $z^{2} + \overset{z}{ˉ}^{2}$ as if it were $z \overset{z}{ˉ} = ∣ z ∣^{2} = x^{2} + y^{2}$ . That gives $x^{2} + y^{2} = 1$ and the wrong answer, circle (D). Another slip is to expand $z^{2}$ as $x^{2} + y^{2}$ , forgetting that $(i y)^{2} = - y^{2}$ . Squaring flips the sign of $y^{2}$ , and that sign is what makes the curve a hyperbola and not a circle or an ellipse.

From the GATE 2024 IN paper. Every question of this paper · Download the full paper · All 24 Complex analysis questions, with notes

GATE 2024 IN Q.11: solution and answer

Answer and solution

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