Wave equations - how to get a real solution from imaginary roots Im trying to follow the derivation on how to solve Laplace equation used in my fluid dynamics course. We are trying to solve for the velocity potential in potential theory. So far we have this:
$$\Phi=F_1(z)\bigl(b_1 e^{i(\omega t+kx)}+b_2 e^{i(\omega t-kx)}\bigr)$$
Here $F_1$ is just a function in the coordinate $z$, and $b_1$ and $b_2$ are arbitrary constants. My teacher now says,

We are only interested in waves travelling to the right in the positive x-direction. We therefore obtain:
  $$F_1 b_2 \sin(\omega t-kx)$$

How can he say that? It seems like he only takes one of the complex solutions but is he allowed to just take the imaginary part of that? Isn't he suppose to simplify the the complex part away?
 A: The Laplace equation $\nabla^2 \psi = 0$ is a linear differential equation. Now note that if $\phi$ is real, then so is $\nabla^2 \phi$. Moreover, by the linearity of the equation, if $\phi$ is real, then $i\phi$ is pure imaginary, and so is $\nabla^2(i\phi) = i \nabla^2(\phi)$.
Okay, back to your situation. Let's say the solution is $\phi_1 + i\phi_2$ for real $\phi_1$ and $\phi_2$. Then the first term gives a real contribution to the RHS, and the second gives a totally imaginary one. Since we know the RHS has to be zero (i.e. both zero real and zero imaginary part), both of these contributions must be zero, which means both $\phi_1$ and $\phi_2$ themselves solve the equation. So we can just take $\phi_1$.
The overarching idea here is that the real and imaginary parts are independent, so you can separate them whenever you want. This is not true for any differential equation that contains complex numbers, like Schrodinger's, but it does work for every other linear differential equation, so you'll see this trick a lot. Unfortunately since it's such a common trick, sometimes people don't bother to explain it.
