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Problem. Find all meromorphic functions $f: C \to C$ s.t. $|f(z)|=1$ wherever $|z|=1$.

$f(z)$ should be like $f=g(z)/h(z)$, where $g(z)$, $h(z)$ are holomorphic functions. I know that if $f(z)$ is also meromorphic at infinity, then it is easy to conclude that $f(z) = g(z)/h(z)$, where $g(z)$, $h(z)$ are polynomials. But now this condition is not satisfied, so I was stuck.

Thanks for opinion.

Sangchul Lee
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user771160
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1 Answers1

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$f$ has a finite number of zeros and poles on $C$; moreover, there is an open set $O$ containing $C$ in which there are no other poles or zeros. Let $Z$ be the set zeros and $P$ the set of poles of $f$ in $C$ (both counted with multiplicity.) Let $$ g(z)=\Bigl(\prod_{p\in P}\frac{z-p}{1-\bar p\,z}\Bigl)\Bigl(\prod_{\zeta\in Z}\frac{z-\zeta}{1-\bar \zeta\,z}\Bigr)^{-1}f(z). $$ Then $g$ is holomorphic in $=$ (after removing the removable singularities at the poles and zeros of $f$), has no zeros in $O$ and $|g(z)|=1$ if $|z|=1$. It follows that $g$ is constant, and $$ f(z)=a\,\Bigl(\prod_{p\in P}\frac{z-p}{1-\bar p\,z}\Bigl)^{-1}\Bigl(\prod_{\zeta\in Z}\frac{z-\zeta}{1-\bar \zeta\,z}\Bigr) $$ where $a$ is a constant with $|a|=1$.

Julián Aguirre
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  • Wow, the construction of g is really fantastic! It successfully removes all the poles and zeros of f, as well as keeping the norm at the boundary of the circle! That work out smoothly! – user771160 Mar 30 '17 at 02:06