Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I saw the following exercise:

If $f:\mathbb{C}\rightarrow\mathbb{C}$ is an entire, non-constant function with only finitely many zeros, then either $|f(z)|\rightarrow \infty$ for $|z|\rightarrow\infty$ or there is a sequence of points $z_n$ such that $|z_n|\rightarrow\infty$ and $f(z_n)\rightarrow 0$.

I thought a bit about this exercise and of course $f$ has to be unbounded because of Liouville's Theorem. But if I assume, that there is a unbounded sequence $z_n$ for which $f(z_n)\rightarrow \infty$ does not hold, how can I conclude, that there has to be a sequence such that $f(z_n)$ goes to zero?

Thanks for hints!

share|improve this question
1  
factor out the zeros and look at the reciprocal –  mike May 1 '12 at 16:39

2 Answers 2

up vote 4 down vote accepted

To add on to what Makuasi stated an entire function has a pole of order $n$ at $\infty$ iff the function is a polynomial of order $n$. So if $f$ does not have a pole at $\infty$ $f$ is either bounded or has an essential singularity at $\infty$. $f$ cannot be bounded by Liouville's Thm. So $f$ must have an essential singularity at $\infty$. By Casorati–Weierstrass theorem there is a sequence, $(z_{n})$, such that $|z_{n}| \rightarrow \infty$, and $f(z_{n})\rightarrow 0$

share|improve this answer
    
thank you Myke, i was just about to write that. :) –  El Angel Exterminador May 1 '12 at 17:21
    
Thank you very much, that helped a lot! –  Fabmor May 3 '12 at 8:53

Hint: An entire function $f(z)$ has a pole of order $n$ at $\infty$ iff $f(z)$ is a polynomial of degree $n$ and $f(z)$ is a trancendental entire function then there exist a sequence $z_n$ such that $|z_n|\rightarrow\infty$ for which $f(z_n)\rightarrow\infty$, Let $f(z)=\sum_{k=0}^{\infty}a_kz^k$ be an entire function and have a pole of order $n$ at $\infty$. if we define $g(z)=f(1/z)$, then $g(z)$ has a pole of order $n$ at the origin , It follows that $z^nf(z)$ is bounded near $0$ i.e $z^{-n}f(z)$ is bounded near $\infty$. That is $f(z)$ is entire such that $f(z)\le M|z|^n$ for $|z|>R$. Can you prove from here that $f(z)$ is a polynomial of degree $n$? the converse part is very trivial.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.