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Entire function


Prove that if f :\mathbb{C} \to \mathbb{C} is an entire function and |z|=1 \Rightarrow |f(z)|=1 then f is constant or there exists c \in \mathbb{C} with |c|=1 and n some positive integer such that f(z)=cz^n.

Solution: f cannot pe identically 0, therefore, the set of zeros of f is a discrete set ( formed by isolated points ). Define g(z)=\displaystyle \frac{1}{\overline{f(\frac{1}{\bar{z}}})} on the set \Omega where \frac{1}{\bar{z}} is never a zero for f. Check by definition that g is also holomorphic on \Omega. Therefore, the equality g(z)=f(z),\ |z|=1 implies f(z)=g(z) on $\Omega$.

Suppose f has a zero z_0 \neq 0. Then f is bounded in a punctured disk around \frac{1}{\bar{z_0}}, which should be a pole for g. But in a sufficiently small punctured disk around \frac{1}{\bar{z_0}} f=g, and thus g is bounded in a neighbourhood of this point, and the singularity is not a pole, but is removable. Contradiction.

Therefore, the only point where f can be 0 is at z=0. Suppose now that f(z)=z^k h(z), where h(0)\neq 0. Then h satisfies the same conditions as f, but is never 0. Defining an analogue g-function for h also, we see that h(\mathbb{C})\subset h(\mathbb{D}) which is bounded. By Liouville’s Theorem, h is constant. Therefore f(z)=c z^k, \ k \geq 0, \ |c|=1.

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  1. April 29, 2010 at 4:57 pm

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