Let z != 1, which module is 1. Prove that z is presentable in the following form:
$$ z =\begin{align}
\frac{1 + ti}{1 - ti}
\end{align}$$
where t is a real number
So, im guessing i have to multiply$$\begin{align} \frac{1 + ti}{1 - ti} with: 1-ti \end{align}$$
to get
$$ z =\begin{align} \frac{1^2 - t^2i^2}{(1 - ti)^2} \end{align}$$
and i also know that $$ i = \sqrt{-1} \begin{align} \ \end{align}$$
But what about the t? How can i find it?
Notice that $$|1\pm ti|=\sqrt{1+t^2}$$ hence $$\left|\frac{1+ti}{1-ti}\right|=1$$ Now we should verify that the map $$\Bbb R\rightarrow\{z\in\Bbb C\;|\; |z|=1\},\quad t\mapsto \frac{1+ti}{1-ti}$$ is surjective, in fact $$\frac{1+ti}{1-ti}=\frac1{1+t^2}\left(1-t^2+2it\right)=\cos\theta+i\sin\theta$$ with $$\cos\theta=\frac{1-t^2}{1+t^2}\;;\; \sin\theta=\frac{2t}{1+t^2}\iff t=\tan\left(\frac\theta2\right)$$