If z=√32+(i)/(2),i=-1, then (z^201-i)^8 is equal to

Question

If $z=\frac{\sqrt{3}}{2}+\frac{i}{2},i=\sqrt{-1}$, then ${({z}^{201}-i)}^{8}$ is equal to

Accepted Answer

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

Convert into Euler form $z=r{e}^{i	heta }$ where $r=|z|$ and $	heta =arg(z)={	an }^{-1}(\frac{y}{x})$ if $z=x+iy$.

$r=|z|=\sqrt{{(\frac{\sqrt{3}}{2})}^{2}+{(\frac{1}{2})}^{2}}=\sqrt{\frac{3}{4}+\frac{1}{4}}=1$

Since $z$ lies in the first quadrant in the complex plane,

$\begin{gathered} 	heta ={	an }^{-1}(\frac{1/2}{\sqrt{3}/2})=\frac{\pi }{6} \ z={e}^{i\frac{\pi }{6}} \end{gathered}$

To calculate ${({z}^{201}-i)}^{8}={({({e}^{i\frac{\pi }{6}})}^{201}-i)}^{8}$

$={({e}^{i\frac{201\pi }{6}}-i)}^{8}$

Using De Moivre's theorem, ${e}^{i	heta }=\cos 	heta +i\sin 	heta $

$\begin{gathered} {({z}^{201}-i)}^{8}={(\cos (\frac{201\pi }{6})+i\sin (\frac{201\pi }{6})-i)}^{8} \ ={(\cos (\frac{67\pi }{2})+i\sin (\frac{67\pi }{2})-i)}^{8} \ =(0+i(-1)-i{)}^{8} \ =(-2i{)}^{8}={2}^{8}(i{)}^{8} \end{gathered}$

Since $i=\sqrt{-1}\implies {i}^{2}=-1\implies {i}^{4}=1\implies {i}^{8}=1$

$	herefore $ ${({z}^{201}-i)}^{8}={2}^{8}=256$.

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