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

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Solution

Let $z = x + i y$
$\Rightarrow \frac{z^{2}}{z - 1} = \frac{(x + i y)^{2}}{x + i y - 1} = \frac{x^{2} - y^{2} + 2 x y i}{x - 1 + i y} \times \frac{x - 1 - i y}{x - 1 - i y}$
$= \frac{(x - 1 - i y) (x^{2} - y^{2} + 2 x y i)}{(x - 1)^{2} + y^{2}}$
$= \frac{x^{3} - x y^{2} + 2 x 62 y i - x^{2} + y^{2} - 2 x y i - x^{2} y i + y^{3} i + 2 x y^{2}}{(x - 1)^{2} + y^{2}}$
$= \frac{x^{3} - x^{2} + y^{2} + x y^{2} + i (2 x^{2} y - 2 x y - x^{2} y + y^{3})}{(x - 1)^{2} y^{2}}$

Given that $\frac{z^{2}}{z - 1}$ is real
Imaginary part= $0$
$\Rightarrow \frac{2 x^{2} y - 2 x y - x^{2} y + y^{3}}{(x - 1)^{2} + y^{2}} = 0$
$\Rightarrow x^{2} y - 2 x y + y^{3} = 0$
$\Rightarrow y (x^{2} - 2 x + y^{2}) = 0$
$\Rightarrow y = 0$ or $x^{2} + y^{2} - 2 x = 0$

$∴$ $z$ lies on $x$ axis or on the circle $x^{2} + y^{2} - 2 x = 0$ .

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If z≠1 and z2z−1 is real, then the point represented by the complex number z lies?

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