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Let $f(z)$ be a function on the unit disk $\mathbb{D}$ which is meromorphic on $\mathbb{D}$ with only one simple pole at $z = 1/2$ and which is continuous up to $∂\mathbb{D}$ and $|f(z)| ≡ 1$ along $∂\mathbb{D}.$ Write down all functions described by the above.

This is one over several parts of a question from an old qual. I want to say there are no such functions, but I do not know how to go about this. And suggestions? Thanks. I tried using shwartz?

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    $\begingroup$ Multiply by ${z-1/2\over 1-z/2}$ which also fixes the boundary disk and you have a function which is analytic on the disk and fixes the boundary. You then have a function analytic on $\mathbb{D}$ preserving $\partial\mathbb{D}$. Use the inversion across the boundary $z\mapsto z^*={1\over\overline{z}}$ to produce a meromorphic continuation to all of $\mathbb{C}\cup\{\infty\}$ and use the classification of these via Blaschke products. $\endgroup$ Jun 29 '14 at 18:35
  • $\begingroup$ @AdamHughes Why are you suggesting analytic continuation? The problem never said to analytically continue anything. In fact, $f(z)$ is only defined on $\overline{\mathbb{D}}$ and the point of the problem was to characterize what $f(z)$ looks like while it is defined on $\overline{\mathbb{D}}$. I agree that in the end $f(z)$ will have an analytic continuation, but that is unnecessary to the problem. $\endgroup$ Jun 29 '14 at 22:27
  • $\begingroup$ Because it makes it easy to see the form of the answer from the big machinery, making it easy to get at the answer. Since the function is meromorphic you may as well extend it if it makes the problem easier. One can prove Gauß' two squares theorem without the Gaußian integers as well, but we use them for simpler proofs. $\endgroup$ Jun 29 '14 at 23:21
  • $\begingroup$ @AdamHughes I agree with your statements, I am just a little concerned about the ordering of your reasoning. You say to product $f(z)$ with a Blaschke factor to obtain a function that is analytic on $\mathbb{D}$ and continuously preserving $\partial \mathbb{D}$. How then do you conclude that you can analytically continue this function to a meromorphic function on all of $\mathbb{C}\cup\{\infty\}$? After all, isn't that the corollary of Blaschke's main result. In other words, it sounds like you are suggesting to use the result that the OP is trying to prove a special case of. $\endgroup$ Jun 30 '14 at 2:16
  • $\begingroup$ The inversion map is easily checked to preserve the boundary, $\partial\overline{\mathbb{D}}$, and the identity theorem is very well established early in a first course in complex variables. Also, the OP said this is from a prelim, and usually the rules there are: "unless stated otherwise, use anything from class to get it done," so I opted for the nuclear option which is good for saving time on those tests and getting on with more problems. If it were first-principles homework I might have opted for a more grass-roots approach. $\endgroup$ Jun 30 '14 at 3:32
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First, we discuss the zeros of $f(z)$. Because we have continuity on the boundary, there can be no accumulation point of zeros at the boundary. Next, because $1/2$ is a simple pole there can be no accumulation point of zeros at $1/2$ either. Thus we conclude that $f(z)$ must have a finite number of zeros (since if they accumulated to a point strictly in $\mathbb{D}-\{1/2\}$, then $f(z)$ would be the zero function). Let $$z_1,\ldots,z_n$$ be the zeros of $f(z)$.

We now show that there exists an $\alpha\in\mathbb{C}$, $|\alpha|=1$, such that $$f(z)=\alpha\cdot \frac{z-2}{1-2z}\cdot\frac{z-z_1}{1-\overline{z_1}z}\cdots\frac{z-z_n}{z-\overline{z_n}z}. $$

Proof. Consider the function, $$g(z):=f(z)\frac{1-2z}{z-2}\cdot\frac{1-\overline{z_1}z}{z-z_1}\cdots\frac{1-\overline{z_n}z}{z-z_n}.$$ We notice, several important facts, $g(z)$ is continuous on $\partial \mathbb{D}$, with $|g(z)|=1$ for $|z|=1$, $g(z)$ is analytic on all of $\mathbb{D}$ (we removed the one singularity that $f(z)$ had), and most importantly, $g(z)$ is not zero in all of $\mathbb{D}$. Hence, neither $g(z)$ nor $\frac{1}{g(z)}$ achieve a maximum on $\mathbb{D}$ (maximum modulus principle). Thus, $g(z)$ must be a constant. But by continuity on the boundry, $|g(z)|=1$, hence there is an $|\alpha|=1$, such that $g(z)=\alpha$, which is what we wished to show.

In general, $$\frac{z-w}{1-\overline{w}z}$$ are called Blaschke products. They typically are used to categorize functions that map the unit circle to the unit circle, since they themselves map the unit circle to the unit circle.

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