Complex Numbers – Powers of Complex Numbers – JEE Main 23 Jan 2026 Shift 2

Solution:

Convert $z$ to polar form:
$$ z = \cos\frac{\pi}{6} + i\sin\frac{\pi}{6} = e^{i\pi/6} $$

Calculate $z^{201}$ using De Moivre’s Theorem:
$$ z^{201} = \left( e^{i\pi/6} \right)^{201} = e^{i \frac{201\pi}{6}} = e^{i \frac{67\pi}{2}} $$

Simplify the angle:
$$ \frac{67\pi}{2} = 33.5\pi = 32\pi + \frac{3\pi}{2} $$
$$ z^{201} = \cos\left(32\pi + \frac{3\pi}{2}\right) + i\sin\left(32\pi + \frac{3\pi}{2}\right) $$
$$ z^{201} = \cos\frac{3\pi}{2} + i\sin\frac{3\pi}{2} = 0 – i = -i $$

Now evaluate the expression:
$$ (z^{201} – i)^8 = (-i – i)^8 = (-2i)^8 $$

$$ = (-2)^8 \cdot i^8 = 256 \cdot (i^4)^2 = 256 \cdot 1 = 256 $$

Ans. (4)