If |z2 - 1| = |z|2 + 1, then z lies on

|z2 - 1|2 = (|x|2 + 1)2 ⇒ (z2 - 1)(z2 - 1) = |z|4 + 2|z|2 + 1 ⇒ ⇒ R (z) = 0 ⇒ z lies on the imaginary axis.

CircleHard

Question

If |z² - 1| = |z|² + 1, then z lies on

A.the real axis

B.an ellipse

C.a circle

D.the imaginary axis.

|z² - 1|² = (|x|² + 1)² ⇒ (z² - 1)(z² - 1) = |z|⁴ + 2|z|² + 1
⇒
⇒ R (z) = 0 ⇒ z lies on the imaginary axis.

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