# Holomorphic function on the unit disk $f$, show the set $z,w\in \mathbb{C}$ such that $f(z)=f(w)$ is not countable

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.

• I suppose you also want $z \neq w$? – WimC Dec 30 '12 at 19:28
• Yeah, that should probably be in there, thanks. – Derek Allums Dec 30 '12 at 19:35
• 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 $(0,1)$. – Robert Israel Dec 30 '12 at 20:59
• 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. – Arthur Dec 30 '12 at 21:20
• Kudos on the "show your work" part. – Pedro Tamaroff Dec 31 '12 at 3:15

I believe I may have a proof that uses well-known results about holomorphic functions of several complex variables and the Baire Category Theorem.

First, recall that zeros of holomorphic functions of more than one variable are not isolated. This is a consequence of a well known theorem due to Hartogs. Now, suppose that $A$ is countable, say $A=\{(z_1,w_1),(z_2,w_2),\dots\}$. The set $$S:=\{(z,w) : |z|,|w| \leq 1/2 : f(z)=f(w)\}$$ is closed, and we have $$S= \{(z,z):|z| \leq 1/2\} \cup \{(z_1,w_1)\} \cup \{(z_2,w_2)\} \cup \dots.$$ By the Baire Category Theorem, one of the sets in the union has non-empty interior. Clearly it cannot be the diagonal, and so we have that one of the points $(z_j,w_j)$ is isolated in $S$. But this point would be an isolated zero of the holomorphic function $g(z,w):=f(z)-f(w)$, which is a contradiction.

• Seems correct, but it also seems rather overkill to invoke the Hartogs theorem for a proof of a result about a function of one complex variable. – user53153 Dec 31 '12 at 2:49
• +1 Nice answer! But I should have mentioned this in the question: this is from a qualifying exam given after a one semester course in real analysis and analysis of functions of one complex variable. Can you think of a similar proof using more elementary means? I had a feeling it had something to do with what you call $g(z,w)$. – Derek Allums Dec 31 '12 at 17:20
• @unit3000-21 : I've thought about a more elementary proof but I can't find one. I'll keep thinking about it though... Interesting problem! – Malik Younsi Dec 31 '12 at 19:23
• @MalikYounsi Thanks very much. I'll continue to think about it too. – Derek Allums Jan 1 '13 at 19:22

Here's a solution that was pointed out to me by a fellow student:

The claim is that the set is either empty or uncountable, in fact. Assume there is one such pair $(z,w) \in A$ such that $f(z) = f(w)$. But since $\mathbb{C}$ is Hausdorff, we can separate each point by an open set such that the open sets are disjoint: $z\in U, w\in V, U\cap V = \varnothing$.

But by the open mapping theorem, the image contains a ball around $f(z)=f(w)$ which is contained in $f(U) \cap f(V)$. That is, $f(U)$ is open, $f(V)$ is open and they have a nontrivial intersection which must too be open and thus contains a ball $B$ which contains $f(z)=f(w)$.

Now we pick some other point $f(z) \neq \alpha \in B$. Then $f^{-1}(\alpha) \supset \lbrace z_0 \neq z\in U, w_0 \neq w \in V \rbrace$ where $f(z_0) = f(w_0)$. Thus, $f^{-1}(B) \subset A$ where $f^{-1}(B)$ is uncountable.

Thus, if there is one point in $A$, there are uncountably many points in $A$.

• I had thought about this too, but I don't think it works. The problem is that it may happens that your $z,w$ satisfy $|z|=|w|=1/2$. In this case, $f^{-1}(\alpha)$ indeed contains points different than $z,w$, but these points may be outside the closed disk centered at $0$ of radius $1/2$, and so these points do not belong to $A$. See Robert Israel's comment. – Malik Younsi Jan 16 '13 at 13:55