Question

Let $z=x+iy$, where $x$ and $y$ are real. The points $(x,y)$ in the $X-Y$ plane for which $\frac{z+i}{z-i}$ is purely imaginary lie on

• A. A straight line
• B. An ellipse
• C. A hyperbola
• D. A circle

Answer 1

Answer: (D)

A circle

Let $z=x+iy$
$\therefore \frac{z+i}{(z-i)}=\frac{x+iy+i}{x+iy-i}$

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

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

Since, $\frac{z+i}{z-i}$ should be purely imaginary

$\therefore Re\left(\frac{z+i}{z-i}\right)=0$

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

$⟹x^2+y^2-1=0$

$⟹x^2+y^2=1$ which is the equation of a circle

Hence, $(x,y)$ lie on a circle.