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

Answer: 8

Let $z=\cos x+i\sin x,x\in [0,2\pi )$. Then
$1=\left|\frac{z}{\bar{z}}+\frac{\bar{z}}{z}\right|=\frac{|z^2+{\bar{z}}^2|}{|z{|}^2}$
$=\mid \cos 2x+i\sin 2x+\cos 2x-i$ $\sin 2x|=2|\cos 2x\mid$
Hence $\cos 2x=1/2$ or $\cos 2x=-1/2$
If $\cos 2x=1/2$, 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 $\cos 2x=-\frac{1}{2}$, then
Alternatively:
$|z|=1\Rightarrow z=\frac{1}{\bar{z}}$
hence $\left|\frac{z}{\bar{z}}+\frac{\bar{z}}{z}\right|=1;z=e^{i\theta }$
$\left|e^{i2\theta }+e^{-i2\theta }\right|=1$
$|2\cos 2\theta |=1$
$\cos 2\theta =\frac{1}{2}$ 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
$z_k=\cos x_k+i\sin x_k,k=1,2,\dots ,8$