Complex Numbers – Euler form – JEE Main 2025 Shift 2

Solution:

Let $z = e^{i\theta}$ since $|z|=1$.
Then $\bar{z} = e^{-i\theta}$.

Substitute these into the given expression:
$$\frac{z}{\bar{z}} = \frac{e^{i\theta}}{e^{-i\theta}} = e^{i2\theta}$$
$$\frac{\bar{z}}{z} = \frac{e^{-i\theta}}{e^{i\theta}} = e^{-i2\theta}$$

The equation becomes:
$$|e^{i2\theta} + e^{-i2\theta}| = 1$$
Using Euler’s formula $(e^{ix} + e^{-ix} = 2\cos x)$:
$$|2 \cos 2\theta| = 1$$
$$|\cos 2\theta| = \frac{1}{2}$$

This implies:
$$\cos 2\theta = \frac{1}{2} \quad \text{or} \quad \cos 2\theta = -\frac{1}{2}$$

We need to find the number of solutions for $\theta$ in the interval $[0, 2\pi)$.
Consequently, $2\theta$ lies in the interval $[0, 4\pi)$.

Case 1: $\cos 2\theta = \frac{1}{2}$
Reference angle is $\frac{\pi}{3}$.
Solutions in $[0, 4\pi)$:
$$2\theta = \frac{\pi}{3}, \quad 2\pi – \frac{\pi}{3} = \frac{5\pi}{3}, \quad 2\pi + \frac{\pi}{3} = \frac{7\pi}{3}, \quad 4\pi – \frac{\pi}{3} = \frac{11\pi}{3}$$
(4 solutions)

Case 2: $\cos 2\theta = -\frac{1}{2}$
Reference angle is $\frac{\pi}{3}$, solutions in 2nd and 3rd quadrants.
Solutions in $[0, 4\pi)$:
$$2\theta = \pi – \frac{\pi}{3} = \frac{2\pi}{3}, \quad \pi + \frac{\pi}{3} = \frac{4\pi}{3}, \quad 3\pi – \frac{\pi}{3} = \frac{8\pi}{3}, \quad 3\pi + \frac{\pi}{3} = \frac{10\pi}{3}$$
(4 solutions)

Total number of solutions = $4 + 4 = 8$.

Ans. (4)