Is this Harmonic Polynomial Identically Zero? 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!
 A: Even one of those two lines forces $f$ to be identically zero.
Its derivative along the line is zero, so its derivative perpendicular to the line is also zero. By integrating, we can show that $f$ is zero on any line parallel to the original, and therefore everywhere.
A: 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.
