What is the physical significane of Complex Time Evolution in EM Waves? [duplicate]

So, I have been having a hard time understanding why there is even a complex phase for EM waves:

$$\phi=\exp[i\omega t]=\cos(\omega t)+i\sin(\omega t)$$

Don't understand why it is there? Any one have an explanation?

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marked as duplicate by Brandon Enright, tpg2114, John Rennie, jinawee, Emilio PisantyMar 3 '14 at 11:17

The field itself has to be real, of course. The imaginary part is added as a computational aid. Dealing with an exponential, whose derivative is itself, is much easier than dealing with a cosine. So we add the imaginary part, do our computation, then dump the imaginary part of the result. It's all simple when the computations are all linear. Sometimes, though, one is called upon to square a field, as in calculating intensity, and in those cases you have to be a little more careful. The resulting expressions then involve complex conjugation. Bottom line: it's a computational convenience.

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Hmmmm, it does indeed make life easier. But taking the square of a complex function, in our case phase, simply gives unity. Taking the square of the trig. representation of the exponential function does not (or I could be wrong?). –  user1886681 Mar 3 '14 at 3:50
@user1886681 Try expanding $\left(cos (\omega t) + i sin (\omega t)\right)\left(cos (\omega t) - i sin (\omega t)\right)$. –  EtaZetaTheta Mar 3 '14 at 4:33
Right! I totally forgot about the complex conjugate :P lol –  user1886681 Mar 3 '14 at 4:36