# 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!

## 2 Answers

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