The number of complex numbers $z$, satisfying $|z|=1$ and $\left|\frac{z}{\bar{z}}+\frac{\bar{z}}{z}\right|=1$, is :

• A

  4
• Correct answer
  B

  8
• C

  10
• D

  6

Solution

$\begin{aligned} & |z|=1 \\ & \left|\frac{z}{\bar{z}}+\frac{\bar{z}}{z}\right|=1 \\ & \Rightarrow\left|z^2+(\bar{z})^2\right|=1 \end{aligned}$ Let $z=x+i y$ $\begin{aligned} & \Rightarrow\left|(x+i y)^2+(x-i y)^2\right|=1 \\ & \Rightarrow\left|2 x^2-2 y^2\right|=1 \\ & \Rightarrow\left|x^2-y^2\right|=\frac{1}{2} \\ & \Rightarrow x^2-y^2=\frac{ \pm 1}{2} \\ & \text { and } x^2+y^2=1 \end{aligned}$ Case I: $x^2-y^2=\frac{1}{2}$ Case II: $x^2-y^2=-\frac{1}{2}$

Hence, the correct answer is option: 'B'.