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Show that if $f(z)$ is a non-constant entire function , then $g(z)=exp(f(z))$ has essential singularity in $z=\infty$.

This question has been answered in the link Thanks!

My approach: Let $f$ analytic (then $f$ has a series), then $exp(f(z))$ is analytic. Then

$$exp(f(\frac{1}{w}))=\sum_{n=0}^{\infty}{\frac{1}{n!}(f(\frac{1}{w}))^{n}}=\sum_{n=0}^{\infty}{\frac{1}{n!}(\sum_{m=0}^{\infty}{\frac{a_{m}}{w^m}})^{n}}$$

We can see that $exp(f(\frac{1}{w}))$ has essential singularity in $w=0$, since there exist negative coefficients. So, $exp(f(z))$ has essential singularity in $w=0$. Is this right? Thanks!

  • Your argument is not valid. This question has been asked many times. Apply the following to $f(\frac 1 z)$ (with $z_0=0$ https://math.stackexchange.com/questions/127168/essential-singularity – Kavi Rama Murthy Dec 27 '18 at 06:00
  • But why, my argument is not valid?? I prove this is a essential singularity. Thanks! –  Dec 27 '18 at 06:03
  • You have to prove that there are infinitely many non-zero terms in negative powers of $w$ to say that function has an essential singularity. – Kavi Rama Murthy Dec 27 '18 at 06:06
  • @KaviRamaMurthy But it was I proof, my serie has infinitely many terms in negative powers of w. –  Dec 27 '18 at 13:44
  • But you need to show that it has infinitely many NONZERO terms....What if, after opening the brackets all the negative coefficients cancel? – N. S. Dec 27 '18 at 21:22
  • @N.S. But is impossible that the negative coefficientes cancels, after we expand the summation. –  Dec 27 '18 at 22:03
  • @Johnny.c Why? If it is impossible, you should be able to prove it, just try ;) – N. S. Dec 28 '18 at 05:34

1 Answers1

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If the singularity at infinity were removable, then $f$ would be bounded. Then by Liouville's theorem is constant. See this answer.

So it appears we can get that it's a non-removable singularity. But as @N.S. points out, it could still be a pole. To rule that out, see this.