# Show that $|e^{ikx-i\omega t}|^2=1$

Here is my attempt: \begin{align*}|e^{ikx-i\omega t}|^2&=|\left[\cos(kx-\omega t)+i\sin(kx-\omega t)\right]\left[\cos(kx-\omega t)+i\sin(kx-\omega t)\right]|\\&=|\cos^2(kx-\omega t)+2i\sin(kx-\omega t)\cos(kx-\omega t)-\sin^2(kx-\omega t)|\\&=|\cos2(kx-\omega t)+i\sin2(kx-\omega t)|\end{align*}

I know that this is incorrect, since the square power is outside the absolute function. But, since it is mandatory to show efforts for homework style questions I must at least make an attempt. In fact my source indicates that $|e^{ikx-i\omega t}|^2=1$.

Could someone please give me some hints or tips on how I can show that $|e^{ikx-i\omega t}|^2=1$?

Assuming $k,x,\omega,t \in \mathbb{R}\;$ this is just $|e^{iz}|^2\;$ with $z\in \mathbb{R},\;$ it is $=1$ because the complex number $e^{iz}$ is on the unit circle. Here with some more detail (where * denotes the complex conjugate): $$|e^{iz}|^2 = e^{iz} (e^{iz})^{*} = e^{iz} e^{-iz}=$$ $$\Big(\cos z + i \sin z\Big)\Big(\cos(-z) + i \sin(-z)\Big)=$$ $$\Big(\cos z + i \sin z\Big)\Big(\cos z - i \sin z\Big)=$$ $$\cos^2 z - i^2 \sin^2 z = \cos^2 z + \sin^2 z = 1$$

Now substitute $z = kx -\omega t\;$ and get

$$\left(\cos(kx-\omega t)+i\sin(kx-\omega t)\right)\left(\cos(kx-\omega t)-i\sin(kx-\omega t)\right)$$ $$=\cos^2(kx-\omega t)+\sin^2(kx-\omega t)=1$$

There's nothing wrong with your calculation (we always have $|z|^2 = \left|z^2\right|$). You could proceed with noting that $|x+\mathrm iy|=\sqrt{x^2+y^2}$ and using that $$(\sin\phi)^2 + (\cos\phi)^2=1$$ to arrive at the desired conclusion.

The following is a much shorter derivation, using only the properties of the exponential function and the absolute value:

If $\phi\in r$. Then $$\left|\mathrm e^{\mathrm i\phi}\right|^2 = \mathrm e^{\mathrm i\phi}\left(\mathrm e^{\mathrm i\phi}\right)^* = \mathrm e^{\mathrm i\phi}\mathrm e^{-\mathrm i\phi} = \mathrm e^{\mathrm i\phi + (- i\phi)} = \mathrm e^0 = 1$$

More generally, $$\left|\mathrm e^{x +\mathrm iy}\right|^2 = \mathrm e^{x +\mathrm iy}\mathrm e^{x -\mathrm iy} = \mathrm e^{2x}$$

$$|e^{i\phi}|^2 = \left(\sqrt{e^{i\phi}e^{-i\phi}}\right)^2 = |e^0| = 1$$

The only way this fails is if $\phi$ is complex, in which case, the statement you sought to prove is false anyway. Sines and Cosines are a diversion.