Trouble figuring out a complex number equation I'm struggling to prove this question, an help is greatly appreciated!
If $a=cis(\pi/5)$, prove that:
$$a^7=-a^2$$
and
$$a^9=-a^4$$
 A: Note $\text{cis}(\theta )= e^{i\theta}$. So,
$$a^7+a^2= e^{i \frac {7\pi}{5}}+e^{i \frac{2\pi}{5}}=e^{i \frac{2\pi}{5}}(e^{i\pi}+1)=0$$
where $e^{i\pi} =-1$. Similarly,
$$a^9+a^4=e^{i \frac{4\pi}{5}}(e^{i\pi}+1)=0$$
A: Rewrite using Euler's formula, which states that $e^{i\theta}=i\sin\theta+\cos\theta.$ The first equation is equivalent to $a^7+a^2=0.$ Using Euler's formula gives$a^7+a^2= (e^{i\frac\pi5})^7+(e^{i\frac\pi5})^2=e^{\pi+2i\frac\pi5}+e^{2i\frac\pi5}=-e^{\pi+2i\frac\pi5}+e^{2i\frac\pi5}=0,$ as required.
Similarly, the second equation can be shown to be true.
$$a^9+a^4=(e^{i\frac\pi5})^9+(e^{i\frac\pi5})^4=e^{\pi+4i\frac\pi5}+e^{4i\frac\pi5}=-e^{4i\frac\pi5}+e^{4i\frac\pi5}=0.$$
Note that this answer can be used to demonstrate an identity; namely, the fact that $(cis(\frac\pi a))^x+(cis(\frac\pi a))^{x-a}=0,$ which is equivalent to saying $(cis(\frac\pi a))^a=-1.$
A: Let's think about it geometrically. $\frac{7 \pi}{5}$ and $\frac{2\pi}{5}$ differ by $\pi$, and so do $\frac{9 \pi}{5}$ and $\frac{4\pi}{5}$.
Geometrically, this means that in the complex plane, $e^{ i\frac{7 \pi}{5}}$ and $e^{i \frac{2\pi}{5}}$ point in opposite directions. Written algebraically, we have $e^{ i\frac{7 \pi}{5}} = - e^{i \frac{2\pi}{5}}$, or $a^7 = - a^2$.
Similarly, $e^{i \frac{9 \pi}{5}}$ and $e^{i \frac{4\pi}{5}}$ point in opposite directions, so $e^{i \frac{9 \pi}{5}} = - e^{i \frac{4\pi}{5}}$, or $a^9 = - a^4$.
