Question

$z=-1+i\sqrt{3}$ and $n$ is a positive integer which is not multiple of $3$, then ${\alpha }^{2n}+2^n{\alpha }^n+2^{2n}$

Answer 1

$z=-1+\sqrt{3}i$

Multiply & divide by 2, we get

$\begin{array}{c}z=2{\left(-\frac{1}{2}+\frac{\sqrt{3}}{2}i\right)}^{}\Rightarrow 2{\left(\cos \frac{2\pi }{3}+\sin \frac{2\pi }{3}i\right)}^{}\\ z=2e^{i\frac{2\pi }{3}}\\ z^{2n}+2^n{z}^n+2^{2n}=2^{2n}\left(e^{\frac{4n\pi i}{2}}+e^{\frac{2n\pi i}{3}}+1\right)\\ =2^{2n}\left[\left(\cos \frac{2\pi n}{3}-i\sin \frac{2\pi n}{3}\right)+\left(\cos \frac{2\pi n}{3}+i\sin \frac{2\pi }{3}+1\right)\right]\\ putn=2\\ \Rightarrow 2^{2n}\left(\frac{-1}{2}\frac{-1}{2}+1\right)\Rightarrow 2^{2n}\left(0\right)=0\end{array}$