Explanation

Given:

$$ z=\frac{\sqrt{3}}{2}+\frac{i}{2} $$

Comparing with $z=\cos\theta+i\sin\theta$:

$$ \cos\theta=\frac{\sqrt{3}}{2},\quad \sin\theta=\frac{1}{2} $$

Hence,

$$ \theta=\frac{\pi}{6} $$

So,

$$ z=\cos\frac{\pi}{6}+i\sin\frac{\pi}{6} $$

Now apply De Moivre’s theorem:

$$ z^{201}=\cos\frac{201\pi}{6}+i\sin\frac{201\pi}{6} $$

Reduce the angle:

$$ \frac{201\pi}{6}=33\pi+\frac{\pi}{2} $$

Since $33$ is an odd integer:

$$ \cos(33\pi+\tfrac{\pi}{2})=-\cos\tfrac{\pi}{2}=0 $$

$$ \sin(33\pi+\tfrac{\pi}{2})=-\sin\tfrac{\pi}{2}=-1 $$

Therefore,

$$ z^{201}=-i $$

Now,

$$ z^{201}-i=-i-i=-2i $$

Raise to power 8:

$$ (-2i)^8=2^8\cdot i^8 $$

$$ =256\cdot1=256 $$

Hence, the required value is 256.

Related JEE Main Questions