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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 $\,|\sin z|\to\infty \,\,if\,\,|z|\to\infty\,$...)2012-06-10

1 Answers 1

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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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    @CameronBuie: Thank you so much for the nice explaining.In this question, does having$a$pole at z =0 rule out the essential singularity. I think it seems clear now. Thanks so much.2012-06-10