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I'm trying to do a question but I have a doubt on holomorphic functions, here is the problem.

Let $A = \{z ∈ \mathbb{C} : \frac{1}{R}< |z| < R\}$. Suppose that $f : A → \mathbb{C}$ is holomorphic and that $|f(z)| = 1 $ if $|z| = 1$. Show that $f(z) = {\left(\overline{f(\bar{z}^{−1})}\right)}^{-1}$ when $ |z| = 1 $ and deduce that this holds for all $z ∈ A$.

Now, my problem is with the final deduction...

I think I need something about the zeroes of f as the right-hand side of the equality could not be defined at some point. Am I right or is there a way to prove that a function with these properties can never be zero in A?

Thank you all.

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p.s. knowing that it is never zero i think the most is done as the rhs is holomorphic as well and you just use the uniqueness theorem as |z|=1 has a limit point in A – Moritzplatz Aug 8 '12 at 14:39
If $|z|=1$ then $(\bar z)^{—1}=z$. – Davide Giraudo Aug 8 '12 at 14:41
Max. modulus principle gives me an upper bound for the modulus of the function doesn't it? What I need is a lower (positive) bound.. I'm not even sure if it is true that such a function has to be non zero, a mobius transformation which maps the unit circle to itself can actually have zeros... – Moritzplatz Aug 8 '12 at 15:15
i think it is true, actually it must be by the uniqueness theorem! – Moritzplatz Aug 8 '12 at 15:23
Yes, you don't need to prove it doesn't have any zeroes, it follows from the uniqueness theorem. – JSchlather Aug 8 '12 at 15:23
up vote 2 down vote accepted

$f(z) \cdot \overline{f(1/\,\overline{z})}$ is well-defined and holomorphic on $A$ and equal to $1$ for $|z|=1$. Then it is constant on $A$, since level sets of non-constant holomorphic functions are discrete.

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thank you, really helpful! I couldn't see the last part so easily (don't really understand the terminology) anyway I did get it using identity theorem! – Moritzplatz Aug 8 '12 at 22:35

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