Let $h$ be a harmonic polynomial that is zero on the lines $Im(z) = 1$ and $Im(z) = -1$.
I know that by the Schwarz Reflection Principle, $h$ must be zero on any line $Im(z) = k$ for $k$ odd, but this fact might not be necessary.
I'm pretty sure that $h$ must be identically zero, but how do I prove this statement?
I've been assuming that a harmonic polynomial is simply an element of $\Bbb R[x,y]$ whose Laplacian is zero, but I've come across expressions like $f(z) = z^n + \overline z^n$, and I'm not sure how those fit into the picture.
I'm just starting to get into harmonic function theory, so I apologize if my questions are very basic. That said, I'll be very grateful for any pointers that you can provide. Thanks!
You are on the right lines with the Schwarz Reflection Principle. Any $h$ satisfying the conditions would have to be zero on infinitely many horizontal lines $Im(z) = k$, but if you now look at any harmonic polynomial restricted to any perpendicular line $Re(z) = m$, it is just a polynomial function in one variable, so that if can only have finitely many zeros (along that line) unless it is identically zero. Thus your function $h$ must be identically zero on any vertical line, and is hence zero in the whole plane.
As I noted in a comment, the fact that $h$ is a polynomial is needed, since the example $e^x\cos y = Re(e^z)$ is an example of a harmonic function which is zero on infinitely many horizontal lines in the complex plane without being identically zero.