Show $e^{it}$ is bounded by $1$

Question

How do we show that $\lvert e^{it} \rvert = 1$? A video lecture argues that $\lvert e^{it} \rvert = \lvert \cos t + i\sin t \rvert$, by Euler's formula, so ${e^{it}}^2 = {(\cos t + i \sin t )}^2$. Expanding the last expression leaves $\cos^2 t + 2i\sin t\cos t - \sin^2 t$. How is this equal to $1$?

Answer 1

Hint: $$|z|=|x+iy|=\sqrt{x^2+y^2}$$

Remark: Of course in general $(e^{it})^2\neq 1$, what we want to prove is $|e^{it}|^2=1,$ which in turn implies $|e^{it}|=1$

Answer 2

how does $|\cos^2 t - \sin^2 t + i \sin t \cos t| = 1$?

if $z = x+iy, |z| = \sqrt {x^2 + y^2}$

$|\cos 2t + i \sin 2t| = \sqrt{\cos^2 2t + \sin^2 2t} = 1$

You should prove these to yourself:

$|z||w| = |zw|,|z^n| = |z|^n, z\bar z = |z|^2$

All of which would be helpful to show what you have been asked to show.