given a complex number $z$, why does $z^n=1$ imply magnitude of $z$ is 1

Question

Given that $z$ is a conplex number and $n$ is a positive integer, I was wondering why $$z^n=1$$

imply that the magnitude of $z$ is 1. I know that $$z^n= r^n( cis(n \theta) )=r^n e^{n \theta}$$.

Answer 1

Because the modulus of a complex number is multiplicative: $$\lvert z^n\rvert=\lvert z\rvert^n=1\iff \lvert z\rvert=1$$ since the modulus is a positive real number.

Answer 2

When $\exists\space\text{z}\in\mathbb{C}\space\wedge\space\exists\space\text{n}\in\mathbb{R}$:

$$\text{z}^\text{n}=\left(\left|\text{z}\right|e^{\left(\arg\left(\text{z}\right)+2\pi\text{k}\right)i}\right)^\text{n}=\left|\text{z}\right|^\text{n}e^{\text{n}\left(\arg\left(\text{z}\right)+2\pi\text{k}\right)i}=$$ $$\left|\text{z}\right|^\text{n}\cos\left(\text{n}\left(\arg\left(\text{z}\right)+2\pi\text{k}\right)\right)+\left|\text{z}\right|^\text{n}\sin\left(\text{n}\left(\arg\left(\text{z}\right)+2\pi\text{k}\right)\right)i$$

Where $\left|\text{z}\right|$ is the absolute value of $\text{z}$, $\arg\left(\text{z}\right)$ is the complex argument of $\text{z}$ and $\text{k}\in\mathbb{Z}$

So, we will have:

$$\text{z}^\text{n}=1\space\implies\space\begin{cases} \left|\text{z}\right|^\text{n}\cos\left(\text{n}\left(\arg\left(\text{z}\right)+2\pi\text{k}\right)\right)=1\\ \\ \left|\text{z}\right|^\text{n}\sin\left(\text{n}\left(\arg\left(\text{z}\right)+2\pi\text{k}\right)\right)=0 \end{cases}$$

So, when we have a complex number $\text{s}$ and the imaginary part is equal to zero (as stated in the system above for your example) we have:

$$\left|\text{s}\right|=\sqrt{\Re^2\left(\text{s}\right)+\Im^2\left(\text{s}\right)}=\sqrt{\Re^2\left(\text{s}\right)+0^2}=\sqrt{\Re^2\left(\text{s}\right)}=\left|\Re\left(\text{s}\right)\right|$$

And in your example the real part equals $1$:

$$\color{red}{\left|\text{z}\right|^\text{n}\cos\left(\text{n}\left(\arg\left(\text{z}\right)+2\pi\text{k}\right)\right)=1}$$