# 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 $$a^9=-a^4$$

• Try proving that $a^5 = -1$ Then, note that $a^7 = a^2\times a^5$, etc... – JMoravitz Oct 26 '19 at 0:54

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,

$$a^9+a^4=e^{i \frac{4\pi}{5}}(e^{i\pi}+1)=0$$

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 $$(cis(\frac\pi a))^a=-1.$$

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 $$a^9 = - a^4$$.