First slide

Geometry of complex numbers

Question

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

Moderate

A

a circle
B

the imaginary axis
C

the real axis
D

an ellipse

Solution

$|z^{2} - 1| = |z^{2}| + 1$ can be written as $|z^{2} + (- 1)| = |z^{2}| + |- 1|$
   $\Leftrightarrow$ $\frac{z^{2}}{- 1}$ is a non-negative real number.
   $\Leftrightarrow$ $z^{2}$ is a non-positive real number.
   $\Leftrightarrow$ $z$ lies on the imaginary axis.

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