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## Proof: (related to "n-th Roots of Unity")

• By hypothesis, $n\in\mathbb Z$ is a positive integer with $n\ge 2,$ $z\in\mathbb C$ is a complex number, and $z^n=1.$
• By polar coordinates of complex numbers we haver $z=|z|.\exp(i\phi)$ for some real numbers $r,\phi\in\mathbb R$ and $0\le \phi < 2\pi,$ $\pi$ being the number pi.
• Since $z^n=1$, taking the absolute value of complex numbers gives us $|z|^n=1.$
• Thus, $r=1.$
• Therefore, $z^n=(\exp(i\phi))^n=\exp(in\phi)=1.$
• By the geralized Euler’s identity, there is an integer $k\in\mathbb Z$ such that $2k\pi=n\phi.$
• Thus, $\phi=\frac{2k}{n}\pi.$
• Because $0\le \phi < 2\pi$, we have $0\le k < n.$
• Thus, if $z\in\mathbb C$ solves $z^n=1,$ then $z=\zeta_k:=\exp\left(2\pi i\frac{k}{n}\right),$ $k=0,1,\ldots,n-1.$
• Vice versa, if $z=\zeta_k,$ then $z^n=\exp\left(i\frac{2k\pi}{n}\right)=1.$
• Finally, if $m(n)\equiv k(n)$ are congruent $k(n)\equiv m(n)$ modulo $n,$ then $m=k+ln$ for some $l\in\mathbb Z.$ But then
$$\exp\left(i\frac{2m\pi}{n}\right)=\exp\left(i\frac{2(k+ln\pi}{n}\right)=\exp\left(i\frac{2k\pi}{n}\right)\cdot \exp\left(i 2l\pi\right)=\zeta_k\cdot 1.$$
q.e.d

| | | | created: 2019-09-15 12:33:44 | modified: 2019-09-15 12:33:44 | by: bookofproofs

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