JEE Mains · Maths · STD 11 - 4.1 complex nubers
If \( z=\frac{\sqrt{3}}{2}+\frac{i}{2}, i=\sqrt{-1} \), then \( (z^{201}-i)^{8} \) is equal to:
- A -1
- B 0
- C 1
- D 256
Answer & Solution
Correct Answer
(D) 256
Step-by-step Solution
Detailed explanation
\(z=\cos \frac{\pi}{6}+i \sin \frac{\pi}{6}\) \(z^{201}=\cos \left(201 \frac{\pi}{6}\right)+i \sin \left(201 \frac{\pi}{6}\right)=-i\) \(\left(z^{201}-i\right)^8=(-2 i)^8=256\)
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