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We've the following 3 complex numbers

$z_1=2\text{cis}(\frac{\pi}{6})~~$, $z_2=2\text{cis}(\frac{5\pi}{6})~~$, $z_3=2\text{cis}(\frac{-\pi}{2})$

We are asked to show that $z_1^{3n}+z_2^{3n}=2z_3^{3n}$

I have calculated $z_1^{3n}=2^{3n}\text{cis}(\frac{n\pi}{2})~~$, $z_2^{3n}=2^{3n}\text{cis}(\frac{5n\pi}{2})~~$, $z_3^{3n}=2^{3n}\text{cis}(\frac{-3n\pi}{2})$.

I'm stuck on how to proceed further to prove the above equations. Any help really appreciated, Please assist .

Thank you , Arif

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  • $\begingroup$ What is "cis" ? $\endgroup$ – Jean Marie Mar 29 '17 at 6:07
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    $\begingroup$ @JeanMarie This. $\endgroup$ – dxiv Mar 29 '17 at 6:21
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    $\begingroup$ @dxiv Thanks. I understand now the pedagogical interest of this notation... $\endgroup$ – Jean Marie Mar 29 '17 at 6:32
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Hint:

  • $z_1^3 = 2^3 \operatorname{cis}\left(3 \cdot \cfrac{\pi}{6}\right)=8\operatorname{cis}\left(\cfrac{\pi}{2}\right)=8i$

  • $z_2^3 = 2^3 \operatorname{cis}\left(3 \cdot \cfrac{5 \pi}{6}\right)=8\operatorname{cis}\left(\cfrac{5 \pi}{2}\right)=8\operatorname{cis}\left(2\pi+\cfrac{\pi}{2}\right)=8\operatorname{cis}\left(\cfrac{\pi}{2}\right)=8i$

  • $z_3^3 = 2^3 \operatorname{cis}\left(-3 \cdot \cfrac{\pi}{2}\right)=8\operatorname{cis}\left(- 3 \cdot \cfrac{\pi}{2}+ 2 \pi\right)=8\operatorname{cis}\left(\cfrac{\pi}{2}\right)=8i$

Therefore $\;z_1^3=z_2^3=z_3^3\,$.

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