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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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    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.2012-06-10
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    Thanks, J.J. I'd just posted it when I realized that flaw....2012-06-10
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    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.2012-06-10
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    Right you are! Now I see it.2012-06-10
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    You seem to remember it better anyway. :) You don't really need Picard's for this.2012-06-10
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    @preeti: You, yourself, noted that if $f(z)=z$, then $f(1/z)$ has a simple pole at $z=0$, yes? It cannot also have an essential singularity there. That provides a counterexample to (1). Is that what you were asking, or did you mean you don't understand the proof of (4)?2012-06-10
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    @preeti: Yes, it does. If a function $f$ has an isolated singularity at $z=a$, then there are exactly three mutually exclusive possibilities: (a) There is some $b\in\Bbb C$ such that $\lim_{z\to a}f(z)=b$. (b) $\lim_{z\to a}|f(z)|=\infty$. (c) $\lim_{z\to a}|f(z)|$ does not exist. The first case happens iff $a$ is a removable singularity; the second, iff $a$ is a pole; the third, iff $a$ is an essential singularity.2012-06-10
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    Alternatively, $f$ has a removable singularity at $z=a$ iff the singular part of of its Laurent series about $z=a$ is $0$ (that is, the Laurent series is just the Taylor series). $f$ has a pole there iff the singular part is non-0 and has only finitely many non-0 terms. $f$ has an essential singularity there iff the singular part has infinitely many non-0 terms. Again, these are exactly the three mutually exclusive possibilities.2012-06-10
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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