Question

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

• A. $4$
• B. $6$
• C. $8$
• D. More than $8$

Answer 1

The correct option is C $8$

Let $z=\cos x+i\sin x$. Then,
$1=|\frac{z}{\overline{z}}+\frac{\overline{z}}{z}|=\frac{|z^2+\overline{z^2}|}{|z{|}^2}=|{\cos }^{2}x-{\sin }^{2}x+i2\sin x\cos x+{\cos }^{2}x-{\sin }^{2}x-i2\sin x\cos x|=|\cos 2x+i\sin 2x+\cos 2x-i\sin 2x|=|2\cos 2x|\Rightarrow \cos 2x=\pm \frac{1}{2}$
Now,
$\cos 2x=\frac{1}{2}\Rightarrow x_1=\frac{\pi }{6},x_2=\frac{5\pi }{6},x_3=-\frac{\pi }{6},x_4=-\frac{5\pi }{6}[\because x\in (-\pi ,\pi \left]\right]$

$\cos 2x=-\frac{1}{2}\Rightarrow x_5=\frac{\pi }{3},x_6=\frac{2\pi }{3},x_7=-\frac{\pi }{3},x_8=-\frac{2\pi }{3}[\because x\in (-\pi ,\pi \left]\right]$