Show for the complex number $z=\frac{1-t^2+2it}{1+t^2},|z|=1$

Question

Show for the complex number $$z=\frac{1-t^2+2it}{1+t^2},|z|=1$$

for all real values of $t$.

How should I prove it? Hope there's someone to explain this type of question.

Thanks in advance.

Answer 1

Hint:

Notice that $1-t^2+2it=(1+it)^2$. Then use: $\left\lvert \frac{a}{b} \right\rvert=\frac{|a|}{|b|}$ for $a,b\in\mathbb{C}$ and $|x+iy|=\sqrt{x^2+y^2}$ for $x,y\in \mathbb{R}$.

Answer 2

$1-t^2+2it=(1+it)^2$, thus $|1-t^2+2it=|1+it|^2=1+t^2$.

Answer 3

Note that

$$|z|^2=z\bar z=\frac{(t-i)^2(t+i)^2}{(1+t^2)^2}=\frac{(t^2+1)^2}{(1+t^2)^2}=1\implies |z|=1$$