Question

Number of complex numbers $z$ such that $|z|=1$ and $|\frac{z}{\overline{z}}+\frac{\overline{z}}{z}|=1$ is
$(\arg z\in [0,2\pi \left)\right)$

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

Answer 1

The correct option is C $8$

Let $z=\cos x+i\sin x,x\in [0,2\pi )$. Then,
$|\frac{z}{\overline{z}}+\frac{\overline{z}}{z}|=1$
$\Rightarrow \frac{|z^2+{\overline{z}}^2|}{|z{|}^2}=1$
$\Rightarrow |\cos 2x+i\sin 2x+\cos 2x-i\sin 2x|=1$
$\Rightarrow 2|\cos 2x|=1$
or, $\cos 2x=\pm \frac{1}{2}$
Now, $\cos 2x=\frac{1}{2}$
$\Rightarrow x_1=\frac{\pi }{6},x_2=\frac{5\pi }{6},x_3=\frac{7\pi }{6},x_4=\frac{11\pi }{6}$

$\cos 2x=-\frac{1}{2}$
$\Rightarrow x_5=\frac{\pi }{3},x_6=\frac{2\pi }{3},x_7=\frac{4\pi }{3},x_8=\frac{5\pi }{3}$