A non-constant harmonic function must have a zero

Question

I have the question:

If $u$ is a non-constant harmonic function, it must have a zero. Where is $u$ defined from the complex to the real set?

My approach has been as follows:

Since $u$ is non-constant harmonic, it must have a harmonic conjugate $v$ such that $f = u + iv$ is analytic. For a $z$ in a circle, we can use the Cauchy Integral formula to define $f(z)$ and then separate it into real and imaginary parts.

Doing this, we get the formula from the mean value property for harmonic functions. But I am very confused about the mean value property and how it ensures a zero. Since $u$ is non-constant, does it take both negative and positive values and hence by IVP, it must take zero as well? Any pointers on how to proceed?

Answer 1

Suppose $u:\mathbb C\to \mathbb R$ is a nonconstant harmonic function, and that $u$ is never $0.$ Because $u$ is continuous, $u(\mathbb C)$ is a connected subset of $\mathbb R.$ That implies $u(\mathbb C)$ is an interval that doesn't contain $0.$ Thus either $u(\mathbb C)\subset (0,\infty)$ or $u(\mathbb C)\subset (-\infty,0).$ Suppose it's the second case. Let $v$ be a harmonic conjugate of $u$ on $\mathbb C.$ Then $f(z) = e^{u+iv}$ is an entire function such that $|f(z)| = e^{u(z)} \le e^0=1$ everywhere. This implies $f$ is a bounded entire function. By Liouville, $f$ is constant. That implies $u$ is constant, contradiction.

Answer 2

The result is true because of the Little Picard Theorem.

Since $u: \mathbb C \to \mathbb R$ is harmonic and non-constant, there is a non-constant holomorphic function $f:\mathbb C \to \mathbb C$ such that $\text{Re}f = u$. The Little Picard Theorem says that $f(\mathbb C) = \mathbb C$ or $f(\mathbb C) = \mathbb C\setminus \{p\}$ for some $p\in \mathbb C$. So, you can take some point of the form $0+i\eta$ on $f(\mathbb C)$. Therefore, there is $z=x+iy$ such that $f(z) = 0+i\eta$ and from that we conclude $u(z) = 0$.