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JEE Main · 2024 · 30 Jan 2024 · Shift 1 · Complex Numbers

If

z = x + iy, xy \neq = 0

, satisfies the equation

z^{2} + i \overset{z}{ˉ} = 0

, then

z^{2}

is equal to :

(A)9
(B)1✓ Correct
(C)4
(D) $\frac{1}{4}$

Answer: (B)

Solution

From

z^{2} = - i \overset{z}{ˉ}

, taking moduli:

∣ z ∣^{2} = ∣ z ∣

, so

∣ z ∣ (∣ z ∣ - 1) = 0

. Since

x y \neq = 0

,

z \neq = 0

, hence

∣ z ∣ = 1

and

∣ z^{2} ∣ = ∣ z ∣^{2} = 1

.

JEE Main · 2024 · 30 Jan 2024 · Shift 1 · Single correct option

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