Simple exponential $\\|e^{j\omega t}|$

Question

I need help understanding this equation: $\int_0^T |e^{j\omega t}|^2 dt$ = $\int_0^T 1.dt$ = T

0-T is only one period, not all T. $\ e^{j\omega t}$ is a periodic complex exponential and \omega is the angular frequency.

$\ e^{j\omega t} = \cos\omega t + j\sin\omega t$

My question is how $\ |e^{j\omega t}|^2$ got evaluated to 1.

Thank you!

Answer 1

$|sin x+j \cos x|^{2}=\sin^{2} x+\cos^{2} x=1$ for all real $x$ so $|sin x+j \cos x|=1$

Answer 2

With

$e^{ikx} = \cos kx + i\sin kx, \tag 1$

$\overline{e^{ikx}} = \cos kx - i\sin kx, \tag 2$

and

$\vert e^{ikx} \vert^2 = e^{ikx}\overline{e^{ikx}}$ $= ( \cos kx + i\sin kx)( \cos kx - i\sin kx)$ $= \cos^2 kx + \sin^2 kx = 1; \tag 3$

note that $i$ "disappears" when we multiply a complex number by its conjugate. Indeed,

$ (\cos kx + i\sin kx)(\cos kx - i\sin kx)$ $= \cos^2 kx - i \cos kx \sin kx + i \cos kx \sin kx + \sin^2 kx$ $= \cos^2 kx + \sin^2 kx = 1. \tag 4$.