# 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!

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.

• I hadn't thought to look at the vertical lines. You're a genius @Old John! – Open Season Nov 15 '13 at 13:52
• @John Narh - just been around a lot of years and seen quite a few tricks of the trade, and some of them have eventually stuck. – Old John Nov 15 '13 at 13:57

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.

• Why must the derivative perpendicular to the line be zero? – Open Season Nov 15 '13 at 3:42
• Because $\frac{df}{d(iz)} = -i\frac{df}{dz}$. – apt1002 Nov 15 '13 at 3:50
• I'm not entirely convinced yet. We can't assume that $f$ is $\Bbb C$-differentiable – Open Season Nov 15 '13 at 3:53
• @apt1002 Isn't $f(z) = z+\overline z$ harmonic and identically zero along a line? (Which ought to be impossible by your argument?) – Old John Nov 15 '13 at 3:59
• In that case I think I don't understand the question, sorry. I have just looked up "harmonic polynomial" and it is not what I thought it was! – apt1002 Nov 15 '13 at 4:04