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Question

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

A

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

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B

on a circle with centre at the origin

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C

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

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D

on the imaginary axis

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Solution

## The correct option is B on a circle with centre at the originExplanation for the correct option.Step 1 : Find the value of $\frac{{z}^{2}}{z-1}$Let $z=x+iy$, now complex ${z}^{2}$ is defined as:$\begin{array}{rcl}{z}^{2}& =& {\left(x+iy\right)}^{2}\\ & =& {x}^{2}+2xyi+{i}^{2}{y}^{2}\\ & =& {x}^{2}+2xyi-{y}^{2}\left[{i}^{2}=-1\right]\\ & =& \left({x}^{2}-{y}^{2}\right)+2xyi\end{array}$Now, simplify the complex number $\frac{{z}^{2}}{z-1}$.$\begin{array}{rcl}\frac{{z}^{2}}{z-1}& =& \frac{\left({x}^{2}-{y}^{2}\right)+2xyi}{x+iy-1}\\ & =& \frac{\left({x}^{2}-{y}^{2}\right)+2xyi}{\left(x-1\right)+iy}\\ & =& \frac{\left[\left({x}^{2}-{y}^{2}\right)+2xyi\right]\left[\left(x-1\right)-iy\right]}{\left[\left(x-1\right)+iy\right]\left[\left(x-1\right)-iy\right]}\\ & =& \frac{\left({x}^{2}-{y}^{2}\right)\left(x-1\right)-\left({x}^{2}-{y}^{2}\right)iy+2xy\left(x-1\right)i-2x{y}^{2}{i}^{2}}{{\left(x-1\right)}^{2}-{i}^{2}{y}^{2}}\\ & =& \frac{\left({x}^{2}-{y}^{2}\right)\left(x-1\right)-\left({x}^{2}y-{y}^{3}\right)i+\left(2{x}^{2}y-2xy\right)i+2x{y}^{2}}{{\left(x-1\right)}^{2}+{y}^{2}}\left[{i}^{2}=-1\right]\\ & =& \frac{\left({x}^{2}-{y}^{2}\right)\left(x-1\right)+2x{y}^{2}}{{\left(x-1\right)}^{2}+{y}^{2}}+\frac{\left(-{x}^{2}y+{y}^{3}+2{x}^{2}y-2xy\right)}{{\left(x-1\right)}^{2}+{y}^{2}}i\\ & =& \frac{\left({x}^{2}-{y}^{2}\right)\left(x-1\right)+2x{y}^{2}}{{\left(x-1\right)}^{2}+{y}^{2}}+\frac{\left({y}^{3}+{x}^{2}y-2xy\right)}{{\left(x-1\right)}^{2}+{y}^{2}}i\end{array}$Step 2 : Find the locus of complex number $z$.It is given that the complex number $\frac{{z}^{2}}{z-1}$ is real. So its imaginary part is zero. So,$\begin{array}{rcl}& & \frac{\left({y}^{3}+{x}^{2}y-2xy\right)}{{\left(x-1\right)}^{2}+{y}^{2}}\end{array}=0\phantom{\rule{0ex}{0ex}}⇒{y}^{3}+{x}^{2}y-2xy=0\phantom{\rule{0ex}{0ex}}⇒y\left({x}^{2}+{y}^{2}-2x\right)=0$So either $y=0$ which represents the real axis, or,${x}^{2}+{y}^{2}-2x=0$ which represents a circle which passes through origin.Thus, the complex number $z$ lies either on the real axis or on a circle passing through the origin.Hence, the correct option is A.

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