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I'm working on this problem:

Let $f$ be an entire function. Suppose $|f(z)|=1$ if $|z|=1$ and $f$ has only one zero in the unit disk $D_1(0)$. Prove that $f(z)=cz$ for some constant $c$.

proof: I write down what I want to prove:

$(1)$ I would like to prove that $\frac{f(z)}{z}$ is entire.

$(2)$ I would like to prove $\left|\frac{f(z)}{z} \right|$ is bounded.

For $(1)$ I think I must apply Riemann's removable theorem to extend analytically $\frac{f(z)}{z}$ to $\mathbb{C}$. It happens if it is bounded at that singularity. I'm not sure if this holds by our second assumption.

For $(2)$ I observe $ \left| \frac{f(z)}{z} \right|\leq 1\quad \ \forall z \in \partial D_1(0). $

So, finally I could apply Liouville's theorem.

I'll appreciate if someone could help me out.

Thanks.

Gabrielek
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Charles Seife
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1 Answers1

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Let $|a|<1$ the zero of $f$ inside the unit disc (note that by hypothesis it is understood to be simple - otherwise something like $z^2$ would work) and $g(z)=\frac{z-a}{1-\bar a z}$ the corresponding Mobius transform. Obviously $h=\frac{f}{g}$ is holomorphic in the unit disc (and near the boundary outside too for $|az| < 1$), $h$ has no zeroes inside the unit disc and $|h|=1$ on the boundary. This implies by maximum modulus applied to $h, \frac{1}{h}$ that $h$ is a constant $c$ of unit modulus in the unit disc, so $f=cg$ in the unit disc. By analytic continuation it follows that $g$ is entire since $f$ is, hence $a=0$ and $f=cz$ as required. Done!

Conrad
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  • Analytic continuation means that, we can apply Riemann's removable theorem? I'm pretty slow to get it, I'm sorry. – Charles Seife Jan 20 '20 at 16:22
  • We know that $f=cg$ in the unit disc and $f,g$ analytic; analytic continuation means $f=cg$ everywhere they are defined since $f-cg$ is an analytic function zero on the unit disc, hence zero everywhere defined, so in particular, it means that since $f$ is entire, $g$ is (or technically it can be defined to be) entire; but if $a \ne 0$, $g$ has a finite pole so it cannot be entire - hence the conclusion. – Conrad Jan 20 '20 at 16:56
  • I see, I think the main trick is when you define $g(z)$. Thanks so much! – Charles Seife Jan 20 '20 at 18:19
  • No problem; Mobius transforms and finite Blaschke products (finite products of Mobius transforms) are the main tool when dealing with one and then finitely many zeroes of analytic functions in the unit disc - and of course infinite Blaschke products when dealing with infinitely many zeroes but those are somewhat trickier – Conrad Jan 20 '20 at 19:09