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Let $f(z) = e^{\frac{1}{1-\cos z}}$.

I want to show that the Laurent's expansion of $f$ near $z = 0$ has infinitely many positive and negative powers of $z$.

For this, I have shown that $z = 0$ is an essential singularity therefore the Laurent expansion of $f$ will have infinitely many negative powers.

I am not able to show that it has infinitely many positive powers.

I tried using the expansion of $\cos z$. The process is becoming really complicated.

Help, please

Gary
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Gitika
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  • Good work so far! You could try to use the fact that $f(z)$ has infinitely many positive powers in its Laurent expansion about $0$ iff $f(1/z)$ has infinitely many negative powers in its Laurent expansion about $0$. – Izaak van Dongen May 15 '21 at 13:08

1 Answers1

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We use the theorem: if $g(z)$ has a non-removable singularity at $a$, then $e^{g(z)}$ is an essential singularity at $a$.

Infinitely many negative terms: Let $g = \frac{1}{1-\cos{z}}$. When $z = 0$, $g$ has a pole, so $e^g = f$ has an essential singularity.

Infinitely many positive terms: We show $f(1/z)$ has an essential singularity at $0$, where $f(1/z) = \exp(\frac{1}{1 - \cos(1/z)})$. $\frac{1}{1 - \cos(1/z)}$ has an essential singularity at $0$ (limit does not exist). So again $f(1/z)$ has an essential singularity at $0$.

koifish
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  • If $\frac{1}{1-cos\frac{1}{z}}$ has essential singularity at $z = 0$ then why does exponential of that have the same? – Gitika May 15 '21 at 21:10
  • $\frac{z}{1-cosz}$ had a pole at $z =0$ but still exponential of that had an essential singularity. – Gitika May 15 '21 at 21:12
  • N also how does non-existence of limit at $0$ imply essential singularity? – Gitika May 15 '21 at 21:15

  • It's covered by the theorem, but basically if $f$ has an essential singularity at $a$, then the image of $f$ around a neighbourhood of $a$ is dense in $\mathbb{C}$, so the image of $e^f$ around a neighbourhood of $a$ would be dense as well as $e$ is surjective on $\mathbb{C^*}$.

  • I don't see your point here. This doesn't contradict the theorem.

  • If the limit exists, it is either a removable singularity or a pole.

    • – koifish May 16 '21 at 01:06
  • If the limit exists finitely, it is a removable singularity. In a pole, the limit does not exist finitely. Right? This is something I guess I read. – Gitika May 16 '21 at 11:27
  • You wrote $\frac{1}{1-cosz}$ has essential singularity at 0 and therefore $f(\frac{1}{z})$. I didn't get this. – Gitika May 16 '21 at 11:30
  • If it's a pole, the limit goes to $\pm \infty$. It is different from an essential singularity. 2. I wrote $1/(1-\cos z)$ has a pole at $0$, which implies $f(z)$ is essential at $0$, not $f(1/z)$.
    • – koifish May 16 '21 at 12:11
  • I meant ${\frac{1}{1-\cos (\frac{1}{z})}}$. in my last comment. – Gitika May 16 '21 at 12:19
  • This is because $f(1/z) = \exp(\frac{1}{1 - \cos(1/z)})$. $\frac{1}{1 - \cos(1/z)}$ has an essential singularity at $0$, so by the theorem mentioned, $\exp(\frac{1}{1 - \cos(1/z)})$ would have an essential singularity at $0$ as well. – koifish May 16 '21 at 12:37
  • Where has the theorem you mentioned, is talking about essential singularity? – Gitika May 16 '21 at 12:58
  • https://math.stackexchange.com/questions/445160/z-0-non-removable-singularity-of-f-rightarrow-z-0-essential-singularity-of – koifish May 16 '21 at 15:41
  • Okay got it thanks – Gitika May 16 '21 at 19:12