Question

The number of complex number z such that $|\frac{z}{\overline{z}}+\frac{\overline{z}}{z}|=1$ and $\left|z\right|=1$ is

• A. $2$
• B. $4$
• C. $6$
• D. $8$

Answer 1

Answer: (B)

$4$

$|\frac{z}{\overline{z}}+\frac{\overline{z}}{z}|=1|\frac{x+iy}{x-iy}+\frac{x-iy}{x+iy}|=1|\frac{x^2+i(xy+xy)-y^2+x^2-i(xy+xy)+y^2}{x^2+y^2}|=1\frac{2x^2}{x^2+y^2}=1=>2x^2=x^2+y^2=>x^2=y^2⟶\left(1\right)=>|z|=1\left(given\right)=>\sqrt{x^2+y^2}=1=>x^2+y^2=1=>x^2=1-y^2=>2y^2=1=>y=\pm \frac{1}{\sqrt{2}}=>x=\pm \frac{1}{\sqrt{2}}$\\

Therefore $z=x+iyz=\pm \frac{1}{\sqrt{2}}\pm \frac{1}{\sqrt{2}}i$

There are four complex numbers.