Question

If $z$ is a complex number not purely real such that imaginary part of $z-1+\frac{1}{z-1}$ is zero, then locus of $z$ is-

• A. a straight line parallel to x-axis
• B. a circle of radius $1$
• C. a parabola with axis parallel to x-axis
• D. a rectangular hyperbola

Answer 1

Answer: (B)

a circle of radius $1$

$z-1+\frac{1}{z-1}=\overline{z-1}+\frac{1}{\overline{z-1}}$

$\Rightarrow$ $z-\overline{z}+\frac{1}{z-1}-\frac{1}{\overline{z}-1}=0$

$\Rightarrow (z-\overline{z})-\frac{\overline{z}-1-z+1}{(z-1)(\overline{z}-1)}=0$

$\Rightarrow (z-\overline{z})(1-\frac{1}{(z-1)(\overline{z}-1)})=0$

$\Rightarrow$ $z=\overline{z}$ or $(z-1)(\overline{z}-1)=1$

But $z$ is not purely real

$(z-1)(\overline{z}-1)=1$

$\Rightarrow$ $z\overline{z}-z-\overline{z}=0$ which is a circle of unit radius.