Question

If $\frac{z-1}{z+1}$ is purely imaginary then what would be the locus of $z$?

Answer 1

Step 1: Simplify the given expression

Let $z=x+iy$

The given expression is $\frac{z-1}{z+1}$.

Upon substituting $z$ in this equation we get,

$\frac{z-1}{z+1}=\frac{x+iy-1}{x+iy+1}$

$=\frac{\left(x-1\right)+iy}{\left(x+1\right)+iy}$

Upon rationalizing this fraction with $\left(x+1\right)-iy$ we get,

$\frac{z-1}{z+1}=\frac{\left(x-1\right)+iy}{\left(x+1\right)+iy}\times \frac{\left(x+1\right)-iy}{\left(x+1\right)-iy}$

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

$=\frac{x^2-1+y^2+2iy}{{\left(x+1\right)}^2+y^2}$ $\left[\because i^2=-1;\left(a+b\right)\left(a-b\right)=a^2-b^2\right]$

$=\frac{\left(x^2+y^2-1\right)}{{\left(x+1\right)}^2+y^2}+i\frac{2y}{{\left(x+1\right)}^2+y^2}$

Step 2: Determine the locus

It is given that $\frac{z-1}{z+1}$ is purely imaginary.

Therefore the real part of this expression will be zero.

$\therefore \frac{\left(x^2+y^2-1\right)}{{\left(x+1\right)}^2+y^2}=0$

$\Rightarrow$ $x^2+y^2-1=0$

$\Rightarrow$ $x^2+y^2=1$

This is the locus of a circle with a radius of $1$ unit.

Hence, the locus of $z$ is a circle with a radius of $1$ unit.