JEE Main 2025 Jan 23 Shift 2, Mathematics Q17: De Moivre's Theorem And Its Applications

Write $z=e^{i\theta }$ with $\theta \in [0,2\pi )$ since $|z|=1$. Then $\frac{z}{\bar{z}}=e^{i2\theta }$ and $\frac{\bar{z}}{z}=e^{-i2\theta }$.

$\frac{z}{\bar{z}}+\frac{\bar{z}}{z}=e^{i2\theta }+e^{-i2\theta }=2\cos 2\theta$, which is real.

The condition becomes $|2\cos 2\theta |=1$, i.e. $\cos 2\theta =\pm \frac{1}{2}$.

As $\theta$ varies over $[0,2\pi )$, $2\theta$ varies over $[0,4\pi )$, covering two full periods. In each period of $2\pi$, the equation $\cos \varphi =\frac{1}{2}$ has $2$ solutions and $\cos \varphi =-\frac{1}{2}$ has $2$ solutions, total $4$ per period.

Over two periods: $4\times 2=8$ solutions.