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Question: If \(\left| z^{2} - 1 \right| = |z|^{2} + 1,\) then z lies on...

If ∣z2−1∣=∣z∣2+1,\left| z^{2} - 1 \right| = |z|^{2} + 1, then z lies on

A

An ellipse

B

The imaginary axis

C

A circle

D

The real axis

Answer

The imaginary axis

Explanation

Solution

Sol. ∣z2−1∣=∣z∣2+1|z^{2} - 1| = |z|^{2} + 1

⇒ ∣z−1∣2∣z+1∣2=(zzˉ+1)2|z - 1|^{2}|z + 1|^{2} = (z\bar{z} + 1)^{2}

⇒ (z−1)(zˉ−1)(z+1)(zˉ+1)=(zzˉ+1)2(z - 1)(\bar{z} - 1)(z + 1)(\bar{z} + 1) = (z\bar{z} + 1)^{2}⇒z+zˉ=z + \bar{z} =0

∴ z lies on imaginary axis.