If $z=-\dfrac12-\dfrac{\sqrt{3}}{{2}}i$, find the value of $z^4+z^8$.

Solution 1:

If $z=- \frac {1} {2} - \frac {\sqrt{3}} {2} i$, find the value of $z^4+z^8$.

\begin{align} &z=-\frac 1 2 - \frac {\sqrt{3}} 2 i. \\ &\Rightarrow z^3-1=0. \tag A \label A\\ \ \\ &\therefore z^2+z+1=0. \tag B \label B\\ \ \\ &\eqref{A} \Rightarrow z^3=1, z^4+z^8=z+z^2. \\ &\eqref{B} \Rightarrow z^2+z=-1. \\ \ \\ &\therefore z^4+z^8=-1. \end{align}

Solution 2:

Indeed, $z=\cos\left(\frac{4\pi}3\right)+\sin\left(\frac{4\pi}3\right)i$. Therefore\begin{align}z^4&=\cos\left(\frac{16\pi}3\right)+\sin\left(\frac{16\pi}3\right)i\\&=\cos\left(\frac{4\pi}3\right)+\sin\left(\frac{4\pi}3\right)i\\&=z\end{align}and\begin{align}z^8&=(z^4)^2\\&=z^2\\&=\cos\left(\frac{8\pi}3\right)+\sin\left(\frac{8\pi}3\right)i\\&=\cos\left(\frac{2\pi}3\right)+\sin\left(\frac{2\pi}3\right)i\\&=-\frac12+\frac{\sqrt3}2i,\end{align}and so $z^4+z^8=-1$.