If z=√32+i2, then (z^201-i)^8 is equal to

Question

If $z=\dfrac{\sqrt3}{2}+\dfrac{i}{2}$, then $(z^{201}-i)^8$ is equal to

Accepted Answer

Write $z$ in Euler form: $r=|z|=\sqrt{\left(\tfrac{\sqrt3}2\right)^2+\left(\tfrac12\right)^2}=1$, and since $z$ lies in the first quadrant, $\theta=\tan^{-1}\left(\dfrac{1/2}{\sqrt3/2}\right)=\dfrac{\pi}{6}$.

So $z=e^{i\pi/6}$, and
$$z^{201}=e^{i\,201\pi/6}=e^{i\,67\pi/2}.$$

Since $67\pi/2 = 33\pi + \pi/2$ (an odd multiple of $\pi$ plus $\pi/2$), $\cos(67\pi/2)=0,\ \sin(67\pi/2)=-1$, so $z^{201}=-i$.

$$(z^{201}-i)^8=(-i-i)^8=(-2i)^8=2^8\cdot i^8=256\cdot1=256.$$

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