# Countable number of zeros

1. May 4, 2012

### Dustinsfl

Let $f$ be a nonzero meromorphic function on the complex plane. Prove that $f$ has at most a countable number of zeros.

Since $f$ is meromorphic on $\mathbb{C}$, $f$ is holomorphic on $\mathbb{C}$ except for some isolated singularities which are poles. Aslo, $f$ being meromorphic we can write $f$ as $g(z)/h(z)$ both holomorphic with $h\neq 0$.

Now does multiplying through help lead to the conclusion?

So we would have $fh = g$. If so, I am not sure with what to do next.

2. May 4, 2012

### micromass

Staff Emeritus
What about using the identity theorem?? If two holomorphic functions agree on a set with a limit point, then they are equal.

3. May 4, 2012

### Dustinsfl

I understand what you are saying but not sure on how to apply it. Should I start over?

4. May 4, 2012

### Dick

Divide the complex plane up into squares. If there were an uncountable number of zeros then one of the squares must contain an infinite number of zeros. Can you prove that?

5. May 7, 2012

### Dustinsfl

I could probably use the Pigeon Hole Principle to show that but how would that help?

6. May 7, 2012

### Dick

Then use micromass's suggestion. Can you show the zeros must have a limit point?

7. May 7, 2012

### Dustinsfl

Suppose $f$ has uncountably many zeros and divide the complex plane into squares of with edges lengths of 1 centered at $a + bi$ where $a,b\in\mathbb{Z}$. By the Pigeon Hole Principle, one of these squares contains uncountably many zeros. Now that we have infinitely many zeros in a bounded set, we have a limit point. By Liouville's Theorem, $f$ must be constant. Therefore, $f$ has at most countably many zeros.

Last edited: May 7, 2012