First slide

Geometry of complex numbers

Question

If $z \neq 1, \frac{z^{2}}{z - 1}$ is real, then point represented by the complex number $z$ lies

Moderate

A

on circle with centre at the origin.
B

either on the real axis or on a circle not passing through the origin.
C

on the imaginary axis.
D

either on the real axis or on a circle passing through the origin

Solution

As $\frac{z^{2}}{z - 1}$ is real, we get
$\begin{array}{l} \frac{z^{2}}{z - 1} = \frac{{\bar{z}}^{2}}{\bar{z} - 1} \\ \Leftrightarrow z^{2} (\bar{z} - 1) = {\bar{z}}^{2} (z - 1) \\ \Leftrightarrow z \bar{z} (z - \bar{z}) - (z - \bar{z}) (z + \bar{z}) = 0 \\ \Leftrightarrow (z - \bar{z}) (z \bar{z} - z - \bar{z}) = 0 \\ \Rightarrow z - \bar{z} = 0 or z \bar{z} - z - \bar{z} = 0 \end{array}$
$\Rightarrow$ $z$ lies on the real axis or $z$ lies on a circle through the origin.

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