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

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Aicharya · Accepted Answer

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

The number of complex numbers $z$, satisfying $|z|=1$ and…

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