JEE Main 2026 Jan 23 Shift 2, Mathematics Q17: De Moivre's Theorem And Its Applications

JEE Main2026Jan 23, Shift 2Mathematics

If $z = \frac{3}{2} + \frac{i}{2}$ , $i = - 1$ , then $(z^{201} - i)^{8}$ is equal to

Solution

$z = cos \frac{π}{6} + i sin \frac{π}{6}$ , so $z^{201} = cos \frac{201 π}{6} + i sin \frac{201 π}{6} = cos \frac{67 π}{2} + i sin \frac{67 π}{2}$ .

$\frac{67 π}{2} = 33 π + \frac{π}{2}$ . Since $33$ is odd, $cos (33 π + \frac{π}{2}) = 0$ and $sin (33 π + \frac{π}{2}) = - 1$ .

$∴ z^{201} = - i$ , and $z^{201} - i = - 2 i$ .

$(- 2 i)^{8} = 2^{8} \cdot i^{8} = 256 \cdot 1 = 256$ .

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