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Question:

Find all holomorphic functions $f(z)$ on $C \setminus \{0\}$ such that

$$f(1) = 1,\ \ \ \ \ \ \ |f(z)| \le \frac{1}{|z|^3}$$

Attempt at solution:

I've discovered that $f(z) = \dfrac{1}{z^3}$ works. And that other monomials $f(z) =z^n \ (n \ne 3)$ don't work. But can't get a general result using power series expansion of $f$.

Also tried to use various Complex Analysis theorems such as Schwarz Lemma, Maximum Modulus Principle, etc. to rule out cases, but not successful.

As such, any help would be much appreciated. Thank you.

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1 Answer 1

up vote 3 down vote accepted

Hint: Consider $g(z)=z^3f(z)$, and show that it has a holomorphic extension to $\mathbf C$. What can you say about that extension? Conclude.

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Thank you jathd. As g(z) would then map the open unit disk into the unit disk, with g(1) = 1 and g(0) = 0 on the extension you proposed, Schwarz's Lemma would imply that g(z) = cz for some unimodular c. Then again g(1) = 1 gives g(z) = z. So indeed f(z) = z^-3 , the function I found earlier, is the only solution. Thanks again. –  Conan Wong Dec 4 '12 at 1:20
    
@ConanWong Why would you have $g(0)=0$? I was thinking more along the lines that $g$ is holomorphic on $\mathbf C$ and bounded. –  jathd Dec 4 '12 at 1:25
    
Ah yes my mistake. g(z) is holomorphic on C and bounded, so constant by Liouville's theorem. As g(z) = 1, f(z) = z^-3 . Thanks again jathd. –  Conan Wong Dec 4 '12 at 1:48

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