If $\left|z^{2}-1\right|=|z|^{2}+1$, then $z$ lies on

JEE Main · 2004 · 24 Apr 2004 · Offline · Complex Numbers

z^{2} - 1 = ∣ z ∣^{2} + 1

, then

z

lies on

Answer: (D)

Write

z = x + i y

, so that

∣ z ∣^{2} = x^{2} + y^{2}

and

z^{2} - 1 = (x^{2} - y^{2} - 1) + 2 i x y

. The condition

∣ z^{2} - 1∣ = ∣ z ∣^{2} + 1

becomes

(x^{2} - y^{2} - 1)^{2} + 4 x^{2} y^{2} = x^{2} + y^{2} + 1

. Squaring both sides and expanding, the left side is

x^{4} + y^{4} + 2 x^{2} y^{2} - 2 x^{2} + 2 y^{2} + 1

while the right side is

x^{4} + y^{4} + 2 x^{2} y^{2} + 2 x^{2} + 2 y^{2} + 1

. Everything cancels except

- 2 x^{2} = 2 x^{2}

, which forces

x = 0

, so

z

is purely imaginary and lies on the imaginary axis.

JEE Main · 2004 · 24 Apr 2004 · Offline · Single correct option