Let z be a complex number such that $|\mathrm{z}+2|=|\mathrm{z}-2|$ and $\arg \left(\frac{z+3}{z-i}\right)=\frac{\pi}{4}$. Then $|z|^{2}$ is equal to :

JEE Main · 2026 · 04 Apr 2026 · Shift 1 · Complex Numbers

Let z be a complex number such that

∣ z + 2∣ = ∣ z - 2∣

and

ar g (\frac{z + 3}{z - i}) = \frac{π}{4}

. Then

∣ z ∣^{2}

is equal to :

Answer: (A)

∣ z + 2∣ = ∣ z - 2∣

places

z

on the perpendicular bisector of

\pm 2

, namely the imaginary axis, so

z = i y

. Then

\frac{z + 3}{z - i} = \frac{3 + i y}{i ( y - 1 )} = \frac{y}{y - 1} - \frac{3}{y - 1} i

. An argument of

\frac{π}{4}

requires equal and positive real and imaginary parts, so

\frac{y}{y - 1} = \frac{- 3}{y - 1}

, giving

y = - 3

; both parts then equal

\frac{3}{4} > 0

. Thus

z = - 3 i

and

∣ z ∣^{2} = 9

JEE Main · 2026 · 04 Apr 2026 · Shift 1 · Single correct option