Take the wave equation
$$\nabla^2\vec{E} = \frac{1}{c^2} \frac{\partial^2 \vec{E}}{\partial t^2},$$
and let $\vec{E}(\vec{r},t)$ be a solution. Indeed taking the real part $\Re(\vec{E}(\vec{r},t))$ yields the physical significant values.

The initial values at $t = 0$ are $\Re(\vec{E}(\vec{r},0))$ and the problem arises here: this does not give you enough information to predict the future evolution of the field!

For example, consider the 1D case with:
$$\begin{align}
f_1(x,t) &= e^{i(kx-\omega t)}, \\
f_2(x,t) &= \cos(kx - \omega t).
\end{align}$$

Note that $\Re(f_1(x,0)) = \Re(f_2(x,0)) = \cos(kx)$. Therefore the two functions have the same initial values at $t = 0$, yet they evolve differently as $t$ progresses.

To fully specify the initial *state* of a solution, you would need $\Re(\vec{E}(\vec{r},0))$ and ~~one of~~ in addition:

- $\Im(\vec{E}(\vec{r},0))$,
- $\Re(\frac{\partial\vec{E}}{\partial t}(\vec{r},0))$,
- $\Im(\frac{\partial\vec{E}}{\partial t}(\vec{r},0))$. [Added in Edit 2].

In your calculation, the moment you took $\Re(\vec{E}(\vec{r},0))$ and applied Fourier Transform, you've implicitly stipulated $\Im(\vec{E}(\vec{r},0)) = 0$. Therefore you changed the initial state to something else, so of course you'd end up with a different solution.

I will explain to you the derivation of Euler's formula simply.

define

$$
f(x)=\cos(x)+i\sin(x)\\
\partial_xf(x)=-\sin(x)+i\cos(x)=i(\cos(x)+i\sin(x))=if(x)
$$

from this you see that : $f(x)=e^{ix}$.

The reason we keep only the cosine term has nothing to do with the derivation.

We are interested in $\psi(x,t)$. With $\psi(x,t)$ some physical **real** observable, the idea is then to solve the equation for a complex $\psi$ which is easier and at the end of the calculations impose on your function to be real.

As an exemple consider the harmonic oscillator $\partial^2_x \psi +w^2\psi=0$ , the solution for a complex $\psi$ is $\psi=Ae^{iwx}+Be^{-iwx}$. Asking for a real $\psi$ gives $\psi = A \cos(wx+\phi)$ or equivalently $\psi = A \sin(wx+\phi)$.

So the answer is that, you need to solve your equation for a complex function ( which is simpler) and at the end of your calculations remember that your function mus be real.Which in many cases means taking the real part but not always.

## Best Answer

We have

\begin{align} E(r,t) &= E_0 \cos(kr - \omega t - \phi)\\ &=E_0\left(e^{i(kr-\omega t -\phi)} + e^{-i(kr - \omega t + \phi)} \right)\\ &= \tilde{E}_0 e^{i(kr-\omega t)} + \tilde{E}^* e^{-i(kr-\omega t)}\\ &= E^{(+)} + E^{(-)} \end{align}

Note I've set $\tilde{E}_0 = E_0 e^{-i\phi}$.

By (a particular choice of) convention, the first term is called the positive frequency term and the second term is called the negative frequency term.

As has been mentioned, if you are adding waves or performing other linear manipulations you can drop the negative frequency term and just work with the positive one, adding in the corresponding negative frequency term at the end of the calculation to recover a

realfinal answer.If you are confused, I recommend performing the manipulations I have shown above so that you have an expression in terms of complex exponentials rather than sines and cosines. However, instead of dropping the negative rotating part as is often (somewhat mysteriously) recommended in my courses and textbooks, just go ahead and keep it. You now have two terms to drag around instead of one but you will see that it is easier to perform manipulations on the complex expressions instead of sinusoidal expressions. You will also see that whenever you do something to the positive frequency term you basically do the complex conjugate thing to the negative frequency term. Then, if you insist, you can take the intuition you have built to understand that in some circumstances you can drop the negative frequency term so that you have less things to write down.

As a practice example try to answer the following question:

What is the amplitude and phase of the wave which is the sum of the two following waves:

\begin{align} E_1(r,t) &= E_1 \cos(kr-\omega t -\phi_1)\\ E_2(r,t) &= E_2 \sin(kr - \omega t + \phi_2) \end{align}

This can be solved in sinusoidal form using trig identities for sums and differences inside sines and cosines. It can also be solved using complex exponentials as described above. I recommend doing it both ways to see the differences.

Finally, to really put the nail in the coffin of this sinusoidal vs. exponential representation I recommend using the exponential formulas to

derivethe trigonometric identities needed to solve the above exercise. These trig identities can alternatively be derived from geometric considerations and drawing funny triangles and labeling the sides but I have a very hard time doing it that way. Once you get used to it, it is quite simple to prove them using the exponential representation as I hope you'll discover.edit: Let me add a bit more to directly address your question: You ask how we know the complex part won't become real and influence the answer. What you should realize is that when you are working in the complex representation (after having ignored the negative frequency part) the complexity of the expression is actually critical to capture the phase of the wave. What you should in fact be concerned about is "how do we know that the negative frequency part will not develop a positive frequency part and affect the answer?" The answer is that if all of the operations are linear then it is only the coefficients of the exponentials which are affected, never the arguments of the exponentials which contain the phase and frequency terms. However, if you begin multiply waves (say you're mixing or homo/heterodyning signals or looking at other non-linear processes) you will see that terms pop up with different factors up in the exponential than you had originally. In this case I would not recommend dropping negative frequency terms as you could easily miss something.

To summarize: In sinusoidal representation the relevant information is contained in the amplitudes and phases of the sines and cosines and in the complex representation the relevant information is contained in the (complex) amplitude of the coefficients of the positive and negative frequency terms, bearing in mind that the information in the coefficient of the positive frequency term is redundant with the information in the negative frequency coefficient as those two terms are complex conjugates of each other.