Question

If |z² – 1| = |z|² + 1, then show that z lies on imaginary axis.

Answer 1

Let z = x + iy,

Then |z² – 1| = |z|² + 1

⇒ |x² – y² – 1 + i2xy| = |x + iy|² + 1

⇒ (x² – y² – 1)² + 4x²y² = (x² + y² + 1)²

⇒ 4x² = 0

i.e., x = 0

Hence z lies on y-axis.