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JEE Main · 2026 · 23 Jan 2026 · Shift 2 · Complex Numbers

If

z = \frac{3}{2} + \frac{i}{2}, i = - 1

, then

(z^{201} - i)^{8}

is equal to

(A)-1
(B)0
(C)1
(D)256✓ Correct

Answer: (D)

Solution

In polar form

z = cos \frac{π}{6} + i sin \frac{π}{6} = e^{iπ /6}

, so

z^{201} = e^{i 201 π /6} = e^{i 33.5 π}

. Reducing the angle,

33.5 π = 16 (2 π) + \frac{3 π}{2}

, hence

z^{201} = cos \frac{3 π}{2} + i sin \frac{3 π}{2} = - i

. Then

z^{201} - i = - 2 i

, and

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

.

JEE Main · 2026 · 23 Jan 2026 · Shift 2 · Single correct option

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