Question

Let $z=x+iy$ be a complex number. If $\frac{1+\overline{z}}{z}$ is a real number, then

• A. $x=-\frac{1}{2}\text{or}y=-\frac{1}{2}$
• B. $x=-\frac{1}{2}\text{or}y=1$
• C. $x=0\text{or}y=-\frac{1}{2}$
• D. $x=-\frac{1}{2}\text{or}y=0$

Answer 1

The correct option is D $x=-\frac{1}{2}\text{or}y=0$

Given : $z=x+iy$, then
$\overline{z}=x-iy$
Now,
$\frac{1+\overline{z}}{z}=\frac{1+x-iy}{x+iy}=\frac{1+x-iy}{x+iy}\times \frac{x-iy}{x-iy}=\frac{x(1+x)-y^2+i(-2xy-y)}{x^2+y^2}$
As $\frac{1+\overline{z}}{z}$ is a real number, so
$\frac{-2xy-y}{x^2+y^2}=0\Rightarrow y(2x+1)=0\Rightarrow y=0\text{or}x=-\frac{1}{2}$