How to prove that $\frac{1-z}{|1-z|}$ is a root of unity with $z=e^{i\frac{2\pi}{n}}$

Question

Let $z=e^{i\frac{2\pi}{n}}$ with $n\ge 1$

Then prove that $$ Z= \frac{1-z}{|1-z|}$$ is also a root of unity and find its order.

Reminder $x$ is said to be the root of the unity of order $n$ if and only if $n$ is the smallest integer such that $x^n =1$.

As a guess first I tried to compute $Z^n$ (without any reason but since $z^n =1$) but it turn out that $Z^n\neq 1.$

How can one find or guess the order of $Z$. Any tips are welcome

Answer 1

When you see $z-1$, it's usually a good choice to set $z=u^2$, in your case $$ u=\exp\left(\frac{i\pi}{n}\right)=e^{i\alpha} $$ which has $|u|=1$.

Then $z-1=u^2-1=u^2-u\bar{u}=u(u-\bar{u})=2iu\sin\alpha$. Then $$ \frac{z-1}{|z-1|}= \frac{2iu\sin\alpha}{2\lvert\sin\alpha\rvert}=iu $$ because $\alpha=\pi/n$ and $\sin\alpha>0$. Now it should be easy.

A different approach is to square the number $w=(z-1)/|z-1|$, to get $$ w^2=\frac{(z-1)^2}{(z-1)(\bar{z}-1)}=\frac{z-1}{z^{-1}-1}=-z $$ Now it's clear that $$ w^{2n}=(-1)^nz^n $$

Answer 2

Hint:

Using the triangle with vertices in $0, 1,z$ we find that : $$\arg(1-z)=\pi-\left(\frac{\pi-\frac{2\pi}{n}}{2} \right)$$