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Let $f:\mathbb C \to \mathbb C$ be an entire function such that $$\lim _{z\to 0}\left|f\left(\frac{1}{z}\right)\right|=\infty. $$ Then which of the following is true?

  1. $f$ is constant

  2. $f$ can have infinitely many zeros

  3. $f$ can have at most finitely many zeros

  4. $f$ is necessarily nowhere vanishing

My Try:- Let $f(z)=z$, I could eliminate 1,2,4. So, Answer is 3. How do I solve formally? $$\lim _{z\to 0}\left|f\left(\frac{1}{z}\right)\right|=\infty.\iff $$ for $\epsilon>0, \exists \delta: |z|<\delta \implies |f(1/z)|>\epsilon.\iff $ Put $w=1/z.$ for $\epsilon>0, \exists \delta: |1/w|<\delta \implies |f(w)|>\epsilon \iff $ for $\epsilon>0, \exists \delta: |w|>1/ (\delta) \implies |f(w)|>\epsilon$. So, the function has no zero out side the closed disk of radius $\delta$. So, Zeroes may or may not be lie inside the disk of radius $\delta$. How do I proceed further?

Truth_searcher
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1 Answers1

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If the zeros lie in a bounded set and there are infinitely many of them, does that imply the zero set has a limit point? If it does, what does that imply?

Allen Bell
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    You mean, suppose it has infinitely many zeroes inside that disk. Then set of zeroes has a limit point in the closed disk. By Identity theorem, the function must vanish identically inside the disk. Right? – Truth_searcher Nov 08 '19 at 02:59
  • You've got it . – Allen Bell Nov 08 '19 at 03:02
  • What happens entire function vanishes inside a closed disk? – Truth_searcher Nov 08 '19 at 06:05
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    If a function vanishes inside any closed disk (which is what you can conclude from the above), then it is identically 0. Even more, if an entire function vanishes inside a single closed disk, it's identically zero since its zeros have a limit point. – Allen Bell Nov 08 '19 at 17:59