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Here is the problem statement:

Suppose $f$ is a holomorphic function on the unit disk. Show that the set $A=\lbrace (z,w) \in \mathbb{C}^2\;|\; |z|,|w| \leq \frac{1}{2}, z\neq w, f(z)=f(w)\rbrace$ is either finite or uncountably infinite.

I'm pretty stuck, but here are some thoughts:

  1. The bound on the $z,w$ is a ltitle odd. The only thing I take away from it is that $|z+w| \leq |z| + |w| = 1$ so the sum stays in the closure of the disk. But this doesn't seem relevant.

  2. I tried to find a clever way to use the identity principle but I couldn't. For example, suppose to the contrary the set $A$ is countably infinite, then it is pointless to consider the function $g(z) = f(z) - f(w_0)$ for some $w_0 \in A$ since there doesn't need to be countably many $z_0$ matching up with $w_0$ in the sense $f(z_n) = f(w_0)$, only countably many pairs $z,w$ with the same images.

  3. I tried representing $f(z)$ as a power series centered at $z_0=0$: $$ f(z) = \sum_{n=0}^\infty a_nz^n $$ and then considering expressions like $$ f(z_0) - f(w_0) = \sum_{n=0}^\infty a_n(z_0^n - w_0^n) $$ to learn something about the coefficients but this didn't lead anywhere.

I think there must be some slick solution but I can't see it, so I'd prefer hints rather than full solutions right now.

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    I suppose you also want $z \neq w$?2012-12-30
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    Yeah, that should probably be in there, thanks.2012-12-30
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    It's easy to use the Open Mapping Theorem to reduce to the case $|z| = |w| = 1/2$. The problem then becomes one of how many times an analytic image of the circle can intersect itself without crossing. I'm pretty sure $1/2$ could be replaced by anything in $(0,1)$.2012-12-30
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    My intuition says that if $A$ is infinite, then it has an accumulation point somewhere, and you can somehow prove (using holomorphicness of $f$ presumably) that there is a whole neighborhood of that accumulation point contained in $A$, thus un-countably many points. I don't know what to do if the accumulation point happens to be on the diagonal.2012-12-30
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    Kudos on the "show your work" part.2012-12-31

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