Q44 / 75 · JEE Main 2026 (23 January Shift 2) · Mathematics
If $z=\frac{\sqrt{3}}{2}+\frac{i}{2}, i=\sqrt{-1}$, then $\left(z^{201}-i\right)^{8}$ is equal to
- A1
- Correct answerB256
- C-1
- D0
Solution
$z = \frac{\sqrt{3}}{2} + \frac{i}{2} = e^{i\pi/6}$.
$z^{201} = e^{i \cdot 201\pi/6} = e^{i \cdot 33.5\pi} = e^{i(33\pi + \pi/2)}$.
Since $e^{i \cdot 33\pi} = e^{i\pi} = -1$, we get $z^{201} = -1 \cdot e^{i\pi/2} = -i$.
$z^{201} - i = -i - i = -2i$.
$(z^{201} - i)^8 = (-2i)^8 = 2^8 \cdot i^8 = 256 \cdot 1 = 256$.
$z^{201} = e^{i \cdot 201\pi/6} = e^{i \cdot 33.5\pi} = e^{i(33\pi + \pi/2)}$.
Since $e^{i \cdot 33\pi} = e^{i\pi} = -1$, we get $z^{201} = -1 \cdot e^{i\pi/2} = -i$.
$z^{201} - i = -i - i = -2i$.
$(z^{201} - i)^8 = (-2i)^8 = 2^8 \cdot i^8 = 256 \cdot 1 = 256$.
Hence, the correct answer is option: 'B'.