- #1
Dustinsfl
- 2,281
- 5
Let [itex]f[/itex] be a nonzero meromorphic function on the complex plane. Prove that [itex]f[/itex] has at most a countable number of zeros.
Since [itex]f[/itex] is meromorphic on [itex]\mathbb{C}[/itex], [itex]f[/itex] is holomorphic on [itex]\mathbb{C}[/itex] except for some isolated singularities which are poles. Aslo, [itex]f[/itex] being meromorphic we can write [itex]f[/itex] as [itex]g(z)/h(z)[/itex] both holomorphic with [itex]h\neq 0[/itex].
Now does multiplying through help lead to the conclusion?
So we would have [itex]fh = g[/itex]. If so, I am not sure with what to do next.
Since [itex]f[/itex] is meromorphic on [itex]\mathbb{C}[/itex], [itex]f[/itex] is holomorphic on [itex]\mathbb{C}[/itex] except for some isolated singularities which are poles. Aslo, [itex]f[/itex] being meromorphic we can write [itex]f[/itex] as [itex]g(z)/h(z)[/itex] both holomorphic with [itex]h\neq 0[/itex].
Now does multiplying through help lead to the conclusion?
So we would have [itex]fh = g[/itex]. If so, I am not sure with what to do next.