$e^{i(kx-\omega t)}$ as a wave

Question

Consider $u(x,t)=\exp(i(kx-\omega t))$. Then, in case that $\omega$ is a real number, this describes a travelling wave with speed $\omega/k$ moving to the right.

This surely lives on the complex plane since $$ u(x,t)=\cos(kx-\omega t)+i\sin(kx-\omega t). $$

In which way can we call this a "wave"? This surely is not to be thought of as some sinusodial wave I first think of when hearing the word "wave"?

Same question with $e^{i\omega t}$.

Answer 1

Write it like $u(z,t) = \cos(kz-\omega t)\cdot \bar x + \sin(kz - \omega t) \cdot \bar y$.

This precise expression gives you a wave of the following form (the cosine and sine describe how the $x,y$ coordinates of the wave move, as it travels through the $z$ direction):

A more general wave, of the form $u(z,t)=A\cos\phi \cdot \bar x + B \sin(\phi + \varphi)\cdot \bar y$ can be represented as a combination of the blue and red curves below:

Addendum:

Noting that $\bar x = (1,0), \bar y = (0, 1)$, one may write, $u(z,t)=\cos \phi +i\sin \phi = \exp(i\phi)$, where $\phi =kz-\omega t$.

Why do I like the representation I wrote first? Well, you can see that if you have a wave travelling in some direction, we can call that direction $z$, the two remaining perpendicular directions $x,y$ and we have an explicit expression of the periodic behaviour of the wave in it's $x$ and $y$ directions.

An analogous analysis can be done with $\exp i\omega t$, can you see it?

Answer 2

The wave is complex as you wrote and it can be thought of as cork-screwing throughout space and time, but to think of it as a wave you will need to consider the real and imaginary parts separately:

On the left is a plot of the real part of the wave and on the right is it's complex part:

As you can see these look more 'wave like'.

You may think of it as a wave when you consider the real part of the wave in one dimension $$\Re[u(x,t)]=\Re[e^{ikx-i\omega t}]=\cos(kx-\omega t)$$ Selecting $\omega \gt 0$ and $k\gt 0$ will result in a rightward travelling wave, with $k\lt 0$ results in a leftward travelling wave. Plotting a graph for $t=1,2,3,....$ will allow you to see this.