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I have the following homework problem:

What kind of singular point does the function $\frac{1}{\cos(\frac{1}{z})}$ have at $z=0$ ?

What I tried:

We note (visually) that $z_{0}$ is the same type of singularity for both $f,f^{2}$ hence the type of singularity of $f(z)=\frac{1}{\cos(\frac{1}{z})}$ have at $z=0$ is the same type of singularity $f^{2}(z)=\frac{1}{\cos^{2}(\frac{1}{z})}$. We recall $$ \frac{1}{\cos^{2}(z)}=1+\tan^{2}(z) $$

We also note that for any constant $z_{0}\in\mathbb{C}$ the singularity of $f,f+z_{0}$ are the same, this can be proved by noting the Laurent expansions of both functions are the same, up to an additive constant.

It remains to determine the singularity type at the origin of $$ \tan(\frac{1}{z}) $$

This is where I'm stuck, we didn't study what the Taylor series of $\tan(z)$.

I also know that type of singularity $$\cos(\frac{1}{z})$$ have, but I don't know how to connect this with the singularity type of $$\frac{1}{\cos(\frac{1}{z})}$$

Can someone please hint me in the right direction ?

Belgi
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2 Answers2

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The singularity is not an isolated singularity, as $\cos(1/z)$ has a sequence of zeroes approaching $0$ (namely $z_n=1/(n\pi+\pi/2)$). In particular, $0$ cannot be an essential singularity of the function.

Ted Shifrin
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  • I don't think that's right, @Ted: first, the expansion for cosine is valid for any $,z,$ , second we're taking values of $,z,$ s.t. $,|z|;$ is very large, which means $,|\frac1z|=\frac1{|z|},$ is very small, and thus the second expansion is valid. – DonAntonio May 28 '13 at 11:40
  • No, @DonAntonio, the OP asked for the singularity at $z=0$, not at $z=\infty$. We're just so used to doing this analysis at $\infty$. So a Laurent expansion at $0$ won't have any annulus on which to be defined, because of the sequence $z_n\to 0$ of singular points. – Ted Shifrin May 28 '13 at 11:50
  • Hello! I just missed that: true, the OP wants that for $;z=0;$ and I, out of habit, assumed at "the other end". Thanks, I shall erase my answer now. – DonAntonio May 28 '13 at 11:54
  • Thanks for pointing this out. I've edited my answer. – J. J. Sep 19 '13 at 11:01
  • I don't understand your answer. J.J. had this right; $z=0$ is an essential singularity of $\dfrac1{\cos(1/z)}$. – robjohn Sep 19 '13 at 11:39
  • @robjohn The classification of singularities into removable ones, poles and essential singularities are (in almost all books) reserved for isolated singularities. – mrf Sep 19 '13 at 12:02
  • @mrf: hmm. I need to read again. I thought that essential singularities included these functions with no Laurent expansion. I agree that an accumulation point of singularities is neither removable nor a pole :-) – robjohn Sep 19 '13 at 12:09
  • @robjohn What would you call the singularity of $\log z$ at $-1$? (Principal branch) – mrf Sep 19 '13 at 12:18
  • @mrf: I am not alone. – robjohn Sep 19 '13 at 12:19
  • @mrf: it is not a singularity on the proper Riemann Surface. I am sorry for the confusion, I was using "these functions with no Laurent expansion" as a description of $\cos(1/z)$ and it's kind, not as a definition of essential singularity. – robjohn Sep 19 '13 at 12:26
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Notice that neither of the limits $\lim_{z \to 0} \frac{1}{\cos(1/z)}$ and $\lim_{z \to 0} \cos(1/z)$ exist. Therefore the singularity is of essential type.

Edit: The conclusion is wrong since $0$ is an accumulation point of poles. (See Ted Shifrin's answer below.)

J. J.
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