Question

If $\frac{z^2}{(z-1)}$ is always real, then $z$ can lie on

• A. The real axis
• B. The imaginary axis
• C. A circle
• D. A parabola

Answer 1

Answer: (A), (C)

The real axis
A circle

Let
$z=x+iy$

$=\frac{z^2}{z-1}$

$=\frac{x^2-y^2+i2xy}{(x-1)-iy}$

$=\frac{(x^2-y^2+i2xy)(x-1-iy)}{(x-1{)}^2+y^2}$

Considering the imaginary part we get
$\Rightarrow (x^2-y^2)(-y)+2x^{2}y-2xy=0$ .... (The given complex number is always purely real).

$\Rightarrow -x^{2}y+y^3+2x^{2}y-2xy=0$
$\Rightarrow x^{2}y+y^3-2xy=0$
$\Rightarrow y(x^2+y^2-2x)=0$
$y=0$ ... equation of real axis.
and $(x-1{)}^2+y^2=1$ ... equation of a circle centered at $(1,0)$