Why are does the argument ($\omega t + k z$) produce waves in the negative direction? There are a couple of ways to express a plane wave (using imaginary numbers):

*

*$e^{i(\omega t - kx)} = \cos(\omega t-kx) + i\sin(\omega t-kx)$


*$e^{i( -\omega t + kx)} = \cos( -\omega t + kx) + i\sin( -\omega t + kx)$


*$e^{i(-\omega t - kx)} = \cos(-\omega t - kx) + i\sin(-\omega t - kx)$


*$e^{i(\omega t + kx)}= \cos(\omega t + kx) + i\sin(\omega t + kx)$
My understanding is that points 1. and 2. represent waves traveling in the positive $x$-direction, while 3. and 4. represent waves traveling in the negative $x$-direction.
My question is then: Why is that? Why does having the same sign on the time and spatial terms result in a wave going backward? If points 1. and 2. are waves going forward, what is the difference between the two waves (are they equivalent, or do the expressions represent two different waves going in the positive $x$-direction)?
I guess one way to "justify" it would be to plot the waves in python, wolfram alpha, or something similar and see how they behave. However, I hope to get a somewhat deeper understanding of what is going on here. It would be nice to see some plots, but then I would still not understand why having the same sign gives a wave that travels backward and a different sign a wave that travels forward.
Edit: Here is a plot of $\cos(2x-3t)$ from wolfram alpha:

I found it helpful seeing this image together with Davide Dal Bosco's answer.
 A: Allow me to drop the complex notation, as it is useful to do the computations, but at the end usually one takes only the real part to get the result.
A travelling wave can be expressed, as you correctly say as
\begin{equation}
\phi^+(x, t) \propto \cos(\omega t - k x)
\end{equation}
or as
\begin{equation}
\phi^-(x, t) \propto \cos(\omega t + k x)
\end{equation}
Assume that you want to focus on the $x$-position of the maximum of the wave at any given time. This can be of course generalized for a point at any phase in the wave.
The maximum occurs at phase equal to zero ($\cos(0)=1$).
the position of the maximum is at any time for a wave traveling towards the positive direction of the $x$-axis is
\begin{equation}
0 = \omega t - k x_{max} \longrightarrow x_{max}(t) = \frac{\omega}{k} t
\end{equation}
As both $k$ and $\omega$ are positive constants, we see that the position of the maximum is moving towards larger $x$-values with time (as we expect).
On the other hand, for a wave traveling towards the negative direction of the $x$-axis is
\begin{equation}
0 = \omega t + k x_{max} \longrightarrow x_{max}(t) = -\frac{\omega}{k} t
\end{equation}
Once again, as both $k$ and $\omega$ are positive constants, we see that the position of the maximum is decreasing with time.
A: The case you are referring to, $(\omega t+kx)$, is a special case of a more general solution.  In general, the one-dimensional wave equation may be written as
$$f(x-ct)+g(x+ct),$$ where $f$ and $g$ are arbitrary functions and $c$ is the wave speed.  Note that $kc=\omega$, so if $g(x)=kx$, you get $g(x+ct)=\omega t+kx$.
Let us just look at $g$.  As I said, these functions are arbitrary, so you can make them look like anything you want.  For the sake of this discussion, lets use the Gaussian distribution:
$$g(x) = e^{-x^2},$$
where I am not paying any attention to units (you would need to normalize the $x$ coordinate with some length scale).  Given the function, this is what the wave looks like at $t=0$.  The peak occurs at $x=0$.  Since the function does not change with time (just the wave), the peak will occur whenever the argument of $g$ is equal to zero.  Thus, the peak will always occur when $x+ct=0$.  Rearranging this equation, we find that the $x$ position of the peak at time $t$ will be at
$$x=-ct.$$
If $t$ is positive, the peak will be at some negative point.  The larger $t$ is, the farther the peak will be in the negative.  Thus, $g$ represents a left-going wave (assuming left is negative).  There was nothing special about our function; regardless of your definition of $g$, it will still be a left-going wave.
In the example above, I picked out the peak as $x+ct=0$.  In general, I could pick any number, $x+ct=\phi$, where $\phi$ is some real number.  With this definition, $\phi$ is called the phase.  Rearranging as above, I can write the left-going position of a particular phase at a specific time as
$x=\phi - ct$.  Regardless of what $\phi$ is, $x$ will go farther and farther to the left with increasing time.
Note that the same discussion applies to $f$, but with everything moving to the right.  Thus, any solution to the one-dimensional wave equation may be written as a combination of a right-going wave ($f$) and a left-going wave ($g$).  This representation of the solution of the wave equation is called d'Alembert's formula.
