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I'm trying to solve this question:

Let $f$ be analytic in a connected open neighbourhood of the closed unit disk. Assume that $|f(z)|=|z+1|$ on the unit circle $|z|=1$, that $f(1)=2$, and that $f$ has simple zeros at $\pm i/2$ and no other zeros in the disk $|z|<1$. Show that these properties determine $f$ uniquely.

Can someone give me any hint?

My attempt was to use something like the Blaschke product decomposition bt $f$ doesn't necessarily map the disk to itself. I tried skimming through some books but I haven't found anything relevant to solve this question.

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    $\begingroup$ Hint: Let $g(z)=f(z)/(1+z)$. $\endgroup$ – David C. Ullrich Dec 24 '17 at 15:14
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    $\begingroup$ dear @DavidC.Ullrich , I guess one can't write $g(z)$ as a Blaschke product since it's not continuous on the boundary (not defined in $z=-1$). Schwarz's Lemma seems of limited interest here as well. So I really don't know how to sue your hint :( $\endgroup$ – Luigi M Dec 24 '17 at 15:49
  • $\begingroup$ Certainly $g$ is defined at $-1$. You're given that $f\in H(U)$, where $U$ is open and $\overline D\subset U$. And $f(-1)=0$. So there exists $g\in H(U)$ such that $f(z) = (1+z) g(z)$. $\endgroup$ – David C. Ullrich Dec 24 '17 at 15:59
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    $\begingroup$ @DavidC.Ullrich I see, so (I just want to be sure I'm doing everything correct) I identify $g$ with a finite Blaschke product and then use $f(1)=2$ ($g(1)=1$) to find the correct root of unity in order to find first $g$ and then $f$ and therefore conclude that $f$ is uniquely determined. Is that right? $\endgroup$ – Luigi M Dec 24 '17 at 18:27
  • $\begingroup$ I guess - I haven't thought about the actual problem any farther than thinking that $g=f/(1+z)$ was the obvious place to start... $\endgroup$ – David C. Ullrich Dec 24 '17 at 20:31
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According to the given details we can write $f(z)=(z+1)g(z)$ for some $g$ : analytic function in a neighborhood of closed unit disk with $g(\pm i/2)=0$ and $g(1)=1$ and $|g(z)|=1$ on the unit circle. Therefore as you expected $g$ is a finite Blaschke product that vanishes at $\pm i/2.$ Thus $g(z)=c\left(\dfrac{4z^2+1}{z^2+4}\right)$ for some uni-modular constant $c.$ Also evaluating this function at $1,$ we can obtain $c=1.$ Hence $$f(z)=(z+1)\left(\dfrac{4z^2+1}{z^2+4}\right).$$

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