Why is this electromagnetic field a wave?

Question

In D.K.Cheng's Field and wave electromagnetics he states the following:

The phasor electric field intensity for a uniform plane wave propagating in the +$z$-direction is $$mathbf{E}(z)=E_0e^{-jkz} $$
where $E_0$ is a constant, $j$ is the imaginary unit and $k$ is the wavenumber.

My question is, why is this a wave? Isn't a wave supposed to be time-dependent? I tried finding the real-notation of the field.
$$Re(mathbf{E}(z))=E_0cos(kz) $$
I get that when $z$ changes the electromagnetic field becomes a cosine function, but can the wavelike behavior also be coordinate dependent. I thought it should be time-dependent?
Can someone clarify?

Philip Wood · Answer

The time dependent factor, $e^{jomega t}$ has been omitted, to save clutter. Readers are supposed to supply it for themselves. The omission is usually made only when dealing with waves with the same $omega$, or with the same wave at different points $z$ in space. In such cases, $e^{jomega t}$ can be factored out of derivations and put in again at the end, if leaving it out causes worry!

Answered by Philip Wood on January 29, 2021

The Photon · Answer

We hid the time dependence when we said we were using a phasor representation of the field.
A phasor is a complex number $A$ representing a time dependent quantity $|A|cos(omega t+angle{A})$, where $omega$ is the previously defined angular frequency at which the instantaneous value of the quantity varies.

Why is this electromagnetic field a wave?

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