Question

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

• A. $x^2+y^2=1$
• B. $x^2+y^2=4$
• C. $x+y=11$
• D. $2xy=x-y$

Answer 1

The correct option is A $x^2+y^2=1$

Let $z=x+iy$

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

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

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

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

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

Since $\frac{z-1}{z+1}$ is purely imaginary, Re $\left(\frac{z-1}{z+1}\right)=0$

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

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

$x^2+y^2=1$ ; which is a circle of radius $1$ unit.

$z=x+iy$ is a complex number such that $x,y$ lies on the circle centered at origin and radius $1$ unit.