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I have encountered the following problem while reading in complex analysis.

Let $U$ be a domain in $\mathbb{C}$ with $z_0 \in U$. Let $\mathcal{F}$ be the family of analytic functions $f$ in $U$ such that $f(z_0) = -1$ and $f(U) \cap \mathbb{Q}_{\geq 0} = \emptyset$, where $ \mathbb{Q}_{\geq0}$ denotes the set of non-negative rational numbers. Is $\mathcal{F}$ a normal family?

[A normal family of functions is a family such that every sequence of functions from the family has a subsequence which converges uniformly on compact subsets of the domain.]

I know I'm supposed to show my work here, but I don't know how to begin to solve this problem. I know that Montel's Theorem says that if $\mathcal{F}$ is locally uniformly bounded, then it is a normal family, but there doesn't seem to be much to work with to show bounds here. And if it's not a normal family, would I have to produce an explicit sequence which has no uniformly convergent subsequence?

Any guidance would be appreciated. Thanks.

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You do know that analytic functions are open mappings, right? Meaning they map open sets to open sets. So the rational numbers seems like a red herring in this problem. – Harald Hanche-Olsen Jun 26 '12 at 19:30
It's not a red herring. I've left this comment on lots of questions, but the trick is so simple and powerful that I'll do it again. Consider the points $0,1,2$. Since the image of the family omits them, the family omits three points and hence must be normal. – Matt Jun 26 '12 at 19:51
Thanks - I wasn't aware of that version of Montel's Theorem. None of the books I've looked at progressed that far. – ec92 Jun 26 '12 at 23:40

As Harald Hanche-Olsen mensioned above, $f$ avoid nonnegative values, and by taking square root the new family formed by functions $\sqrt{-f}$ take values in the right half plane, then we can use the result of exercise 1 on page 227 in Ahlfors' book, to give the answer of your question.

For the cited exercise, you can prove it like this: use some fractional linear transformation to map the right half plane biholomorphically onto the unit disk, its composite with $f$ forms a (locally) bounded family, hence normal, then every sequence $g_n$ of functions from the new family has a subsequence which converges uniformly on compact subsets of the domain to a function $g$, then you can easily prove that certain fractional linear transformation (in fact, the inverse of the previous one) of $g$ is the corresponding (uniform) limit of the original (sub)sequence $f_n$.

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