If a cosine wave satisfies Maxwell's equations then how does a sine satisfy the equations as well?

Question

Say the real part of
$$tilde{mathbf{E}}(z, t)=tilde{mathbf{E}}_{0} e^{i(k z-omega t)}$$
satisfies all Maxwell's equations. Then how can we say the imaginary part satisfies the equations as well?
Griffiths says in a footnote

Because the real part of E differs from the imaginary part only in the replacement of sine by cosine, if the former obeys Maxwell's equations, so does the latter.

Farcher · Answer

Changing the phase by $frac pi 2$ results in $tilde{mathbf{E}}(z, t)=tilde{mathbf{E}}_{0} e^{i(k z-omega t)}$ becoming $tilde{mathbf{E}}(z, t)=tilde{mathbf{E}}_{0} e^{i(k z-omega t-frac pi 2)}$.
However $e^{-ipi/2}=-i$ so the new solution is $tilde{mathbf{E}}(z, t)=-i,tilde{mathbf{E}}_{0} e^{i(k z-omega t)}$ which is the original solution multiplied by a constant term, $-,i$.
If you take the real part you now get a sine function.

If a cosine wave satisfies Maxwell's equations then how does a sine satisfy the equations as well?

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