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Let $\displaystyle f$ be an entire function such that $$\lim_{|z|\rightarrow \infty} |f(z)| = \infty .$$ Then,

  1. $f(\frac {1}{z})$ has an essential singularity at 0.

  2. $f$ cannot be a polynomial.

  3. $f$ has finitely many zeros.

  4. $f(\frac {1}{z})$ has a pole at 0.

Please suggest which of the options seem correct.

I am thinking that $f$ can be a polynomial and so option (2) does not hold.

Further, if $f(z) = \sin z $ then it has infinitely many zeros... which rules out (3) while for $f(z) = z$ indicates that it has a simple pole at $0$ and option (4) seems correct.

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Your reasoning looks fine to me as long as you can prove every claim made (for example, that $\,|\sin z|\to\infty \,\,if\,\,|z|\to\infty\,$...) –  DonAntonio Jun 10 '12 at 18:36

1 Answer 1

You are correct about $2$, and given that, you should be able to determine whether $1$ is true or not--consider your example $f(z)=z$.

Your example $f(z)=\sin z$ does not meet the given criteria. Note that if there are infinitely many zeros, then the set of zeros is necessarily unbounded, for if not, it has a limit point, and so the function is identically zero, contradicting our assumption that $\lim_{|z|\to\infty}|f(z)|=\infty$. But then we have a sequence $\{z_n\}$ such that $|z_n|\to\infty$ but $f(z_n)=0$ for all $n$, so that once again contradicts our assumption. That takes care of $3$.

For $4$, note that since $|1/z|\to\infty$ as $z\to 0$, then by assumption, $\lim_{|z|\to 0}|f(1/z)|=\infty$, which means that $f(1/z)$ has a pole at $z=0$. (H/T to J.J. for reminding me of that characteristic of poles.)

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Looks good otherwise, but $|e^z|$ does not go to $\infty$ as $z \to \infty$ (consider $z = iy$ where $y \in \mathbb{R}$). For 4 you can see that there is a singularity but it cannot be essential so it's a pole. –  J. J. Jun 10 '12 at 18:46
    
Thanks, J.J. I'd just posted it when I realized that flaw.... –  Cameron Buie Jun 10 '12 at 18:48
    
As to why I think that 4 can't be essential: If it were, then every complex number (except maybe one, Picard's Theorem) would be obtained in any neighbourhood of $0$. But clearly for small enough neighbourhood the obtained numbers have absolute value greater than some constant $M$ (because $f(1/z) \to \infty$ as $z \to 0$), which rules this out. –  J. J. Jun 10 '12 at 18:51
    
Right you are! Now I see it. –  Cameron Buie Jun 10 '12 at 18:53
    
You seem to remember it better anyway. :) You don't really need Picard's for this. –  J. J. Jun 10 '12 at 18:57

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