CircleHard
Question
If |z2 - 1| = |z|2 + 1, then z lies on
Options
A.the real axis
B.an ellipse
C.a circle
D.the imaginary axis.
Solution
|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.
⇒
⇒ R (z) = 0 ⇒ z lies on the imaginary axis.
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