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 $1$ unit
• C. a parabola with axis of symmetry parallel to $x$-axis
• D. a hyperbola

Answer 1

The correct option is B a circle of radius $1$ unit

Imaginary part of $z-1+\frac{1}{z-1}$ is zero.
So,
$z-1+\frac{1}{z-1}=\overline{z-1}+\frac{1}{\overline{z-1}}\Rightarrow z-\overline{z}+\frac{1}{z-1}-\frac{1}{\overline{z}-1}=0\Rightarrow z-\overline{z}+\frac{\overline{z}-1-z+1}{(z-1)(\overline{z}-1)}=0\Rightarrow (z-\overline{z})(1-\frac{1}{(z-1)(\overline{z}-1)})=0\Rightarrow z=\overline{z}\text{or}(z-1)(\overline{z}-1)-1=0$

As $z$ is not purely real, so
$(z-1)(\overline{z}-1)=1$
Assuming $z=x+iy$, then
$(x-1+iy)(x-1-iy)=1\Rightarrow (x-1{)}^2+y^2=1$
The locus is circle whose centre is $(1,0)$ and radius is $1.$