Question

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

• A. 0
• B. 2
• C. 4
• D. 8

Answer 1

Answer: (D)

$|z|=1\Rightarrow z=\cos \theta +i\sin \theta$ for some $\theta \in [0,2\pi )$
Now, $|z|=1\Rightarrow |z{|}^2=1\Rightarrow z\bar{z}=1$.
Thus, $\frac{z}{\bar{z}}+\frac{\bar{z}}{z}=z^2+\frac{1}{z^2}$
$=(\cos \theta +i\sin \theta {)}^2+(\cos \theta -i\sin \theta {)}^2$
$=2\left({\cos }^2\theta +i^2{\sin }^2\theta \right)=2(\cos 2\theta )$

$\therefore \left|\frac{z}{\bar{z}}+\frac{\bar{z}}{z}\right|=1$ $\Rightarrow |\cos 2\theta |=\frac{1}{2}$
$\Rightarrow \cos 2\theta =\pm 1/2$
$\Rightarrow 2\theta =\pi /3,2\pi /3,4\pi /3,5\pi /3,7\pi /3,8\pi /3,10\pi /3,11\pi /3$
$\Rightarrow \theta =\pi /6,\pi /3,2\pi /3.5\pi /6,7\pi /6,4\pi /3,5\pi /3,11\pi /6$
Hence, there are 8 values of $z$