# complex numbers- hurwitz theorem

• November 21st 2010, 08:11 AM
hermanni
complex numbers- hurwitz theorem
Suppose that D is a connected set open set , $f_{n} \in H(D)$ and
$f_{n} -> f$ uniformly on compact subsets of D. If f is nonconstant and
$z \in D$ then there exists N and a sequence $z_{n} -> z$ such that $f_{n} ( z_{n} ) = f(z)$ for all $n \geq N$
Hint: Assume that f(z) = 0. Apply the hurwitz theorem in the disk $D(z, r_{j})$ for a suitable sequence $r_{j}->0$

I didn't even understand the hint.hurwitz theorem states that $f_{n}$
and f have the same number of the zeroes in the disk. Can anyone help, thanx.
• November 22nd 2010, 06:24 PM
Jose27
Quote:

Originally Posted by hermanni
Suppose that D is a connected set open set , $f_{n} \in H(D)$ and
$f_{n} -> f$ uniformly on compact subsets of D. If f is nonconstant and
$z \in D$ then there exists N and a sequence $z_{n} -> z$ such that $f_{n} ( z_{n} ) = f(z)$ for all $n \geq N$
Hint: Assume that f(z) = 0. Apply the hurwitz theorem in the disk $D(z, r_{j})$ for a suitable sequence $r_{j}->0$

I didn't even understand the hint.hurwitz theorem states that $f_{n}$
and f have the same number of the zeroes in the disk. Can anyone help, thanx.

I remember doing this exercise once and using a Cantor's diagonal argument somewhere, but I can't remember why I did all that. Take a look at this and let me know if something's off (It's pretty simple, which makes me wary):

Pick $z_0\in D$ with $f(z_0)=0$. There exists an $r>0$ such that $f(z)\neq f(z_0)=0$ for all $z\in \overline{D}_{r}(z_0)\setminus \{ z_0 \} \subset D$ (the closed punctured disk or radius $r$ and center $z_0$), and by Hurwiz theorem there exists $N$ such that for all $n\geq N$ the functions $f_n$ have the same zeroes as $f$ in the disk, so there exists $z_n\in D_{r}(z_0)$ with $f_n(z_n)=0$. It now suffices to prove that $z_n \rightarrow z_0$, but this follows since $\overline{D}_r(z_0)$ is compact (therefore sequentially compact) and the fact that the sequence $z_n$ can only have one accumulation point by how we picked our disk and the fact that $|f_n(y_n)-f(y)| \rightarrow 0$ whenever $y_n,y\in D$ and $y_n\rightarrow y$.

Check it carefully, maybe I overlooked something.
• November 23rd 2010, 10:34 AM
hermanni
Everything seems OK to me , thank u very much :))