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I do not understand how to use the following information: If $f$ is entire, then

$$\lim _{|z| \rightarrow \infty} \frac{f(z)}{z^2}=2i.$$

So if $f$ is entire, it has a power series around $z_0=0$, so $f(z)=\Sigma_{n=0}^\infty a_nz^n$, and then we get

$$\lim _{|z| \rightarrow \infty} \frac{\Sigma_{n=0}^\infty a_nz^n}{z^2}=2i.$$

How do I continue from here?

It is a part of a question. I just want to know how can I use this info. I don't know how I can manipulate summations, and since it's $|z| \rightarrow \infty$ and not $z \rightarrow \infty$ (which is meaningless), I don't really know what I can do here.

Maybe

$$\lim _{|z| \rightarrow \infty} \Sigma_{n=0}^\infty a_nz^{n-2}=2i,$$ but then what?

Thanks in advance for your assistance!

Harry
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  • Could you post the entire question to give a bit more of the context? – MathematicianByMistake May 30 '15 at 13:14
  • It's hard to tell how you can use the info without knowing what for you want to use it – Hagen von Eitzen May 30 '15 at 13:18
  • Please note that your question has been edited and slightly reformatted. Does this this new version preserve the meaning of the original? If so, please state clearly what are your assumptions, and what you are trying to show. Right now, your question is meaningless. – Alex M. May 30 '15 at 13:38

2 Answers2

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Maybe $\lim _{|z| \rightarrow \infty} \Sigma_{n=0}^\infty a_nz^{n-2}=2i$, but then what?

Then you note that this is only possible when $a_n = 0$ for $n>2$ and $a_2 = 2i$. That is $f$ is a polynomial of degree $2$ with leading coefficient $2i$.

quid
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  • Can you explain further why is that? – Harry May 30 '15 at 13:53
  • You can infer that $|f(z)| \le C |z|^2$ for some constant and the apply http://math.stackexchange.com/questions/143468/entire-function-bounded-by-a-polynomial-is-a-polynomial – quid May 30 '15 at 14:11
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One thing you could do is consider the function $g\colon\mathbb C^\times\to\mathbb C$, $g(z)=z^2f(1/z)$. What happens at $z=0$?

Hagen von Eitzen
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