Question

Find the locus of point z in the Argand plane if $\frac{z-1}{z+1}$ is purely imaginary.

Answer 1

Given,

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

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

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

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

$=\frac{x^2+y^2-1+i2y}{(x+1{)}^2+y^2}$

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

Since given is purely imaginary, $\Rightarrow Re\left[z\right]=0$

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

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

$x^2+y^2=1$

The above equation represents the circle with unit radius.

Therefore z lies on the circle with center origin and radius of 1 unit