Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 ?

share|cite|improve this question
can you mark the right answer as correct one? – Jesse P Francis Apr 6 '15 at 7:06
@JessePFrancis - thanks – Belgi Apr 6 '15 at 8:14
up vote 7 down vote accepted

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.

share|cite|improve this answer
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

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.)

share|cite|improve this answer
Are you sure ? – Belgi May 28 '13 at 10:07
@Belgi: What should that link show? WA, too, claims that the limit does not exist. – J. J. May 28 '13 at 10:08
I missunderstood you, I thought you wrote the limits does exist, sorry – Belgi May 28 '13 at 10:09
Why do you say that your conclusion is wrong? Your reasoning is completely correct. An accumulation of poles indicates an essential singularity. – robjohn Sep 19 '13 at 11:43
Hold on my last comment. There may be a definition conflict. – robjohn Sep 19 '13 at 12:10

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.