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I'm struggling to prove this question, an help is greatly appreciated!

If $a=cis(\pi/5)$, prove that: $$a^7=-a^2$$ and $$a^9=-a^4$$

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

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

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

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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 $a^9 = - a^4$.

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