Question

Number of complex numbers $z$ such that $|z|=1$ and $\left|\frac{z}{\bar{z}}+\frac{\bar{z}}{z}\right|=1$ is

Answer 1

Let $z=\cos x+i \sin x, x \in[0,2 \pi)$. Then
$1=\left|\frac{z}{\bar{z}}+\frac{\bar{z}}{z}\right|=\frac{\left|z^{2}+\bar{z}^{2}\right|}{|z|^{2}}$
$=\mid \cos 2 x+i \sin 2 x+\cos 2 x-i$ $\sin 2 x|=2| \cos 2 x \mid$
Hence $\cos 2 x=1 / 2$ or $\cos 2 x=-1 / 2$
If $\cos 2 x=1 / 2$, then
$x_{1}=\frac{\pi}{6}, x_{2}=\frac{5 \pi}{6}, x_{3}=\frac{7 \pi}{6}, x_{4}=\frac{11 \pi}{6}$
If $\cos 2 x=-\frac{1}{2}$, then
Alternatively:
$|z|=1 \Rightarrow z=\frac{1}{\bar{z}} $
hence $\left|\frac{z}{\bar{z}}+\frac{\bar{z}}{z}\right|=1 ; z=e^{i \theta} $
$\left|e^{i 2 \theta}+e^{-i 2 \theta}\right|=1$
$|2 \cos 2 \theta|=1 $
$\cos 2 \theta=\frac{1}{2} $ or $-\frac{1}{2} $
$x_{5}=\frac{\pi}{3}, x_{6}=\frac{2 \pi}{3}, x_{7}=\frac{4 \pi}{3}, x_{8}=\frac{5 \pi}{3}$
Hence there are eight solutions
$z_{k}=\cos x_{k}+i \sin x_{k}, k=1,2, \ldots, 8$