I am to determine the poles of the function

$$f(z)=\frac{1}{\cos(\frac{1}{z})} $$

my texbook sais that this has poles at $$ z_k=(\frac{\pi}{2}+k\pi)^{-1} $$ where $k$ is an integer, and that they are of the order $1!$

but... since $$\cos(z)=\sum_{n=0}^{\infty} \frac{(-1)^nz^{2n}}{2n!}$$

then $$cos(\frac{1}{z})=\sum_{n=0}^{\infty} \frac{(-1)^n}{z^{2n}2n!}=1-\frac{1}{2z^2}-\frac{1}{12z^4}...$$

so inverting that we get $$\frac{1}{\cos(\frac{1}{z})}=\frac{1}{1-\frac{1}{2z^2}-\frac{1}{12z^4}...}$$

and by knowing the series of $$\frac{1}{1-z}=\sum_{n=0}^{\infty} z^n$$ we get

$$\frac{1}{\cos(\frac{1}{z})}=1+(1-\frac{1}{2z^2}-\frac{1}{12z^4}...)+(1-\frac{1}{2z^2}-\frac{1}{12z^4}..)^2+(1-\frac{1}{2z^2}-\frac{1}{12z^4}..)^3+.... $$

how can this have poles of order $1$, aren't these essential poles? if they are (somehow) poles of order $1$, how can I prove it?

  • 2
    $\begingroup$ Something seems to be wrong, since $$\frac1{\frac\pi2+k\pi}=\frac2{k+2k\pi}$$and $\,\cos\,$ doesn't vanish at the above point...Check this. Now, $\;z=0\;$ is a singular point. $\endgroup$ – DonAntonio Jan 19 '18 at 23:10
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    $\begingroup$ There's a typo somewhere either in this problem. Either the function is $1/\cos z$, in which case the poles are the $z_k$ as given, or the function is $1/\cos(1/z)$, in which case the poles are $1/z_k$ as given. You're only going to see things from the Laurent series if you expand around the given point $z_k$, not around $0$. [Note that your power series manipulations do not hold in the whole complex plane. They only hold up to the first singularity as you go away from the origin.] $\endgroup$ – Ted Shifrin Jan 19 '18 at 23:10
  • $\begingroup$ it was a typo, I forgot the -1 power, sorry :) but the problem is basically how to prove that the poles are of order 1 (or of any order for that matter) $\endgroup$ – Alexandar Solženjicin Jan 19 '18 at 23:15
  • $\begingroup$ You have expanded as a Laurent series about the point $z=0$, and found that this is not a pole - it is an essential singularity. This is correct - you do have an essential singularity at $z=0$. To find information about other $z_k$'s using Laurent series, you'd need to do the expansion about $z_k$. These $z_k$'s in particular are poles of order $1$ because these values of $z_k$ are simple zeros of $\cos(1/z)$. $\endgroup$ – John Doe Jan 19 '18 at 23:38
  • $\begingroup$ and why are the zeros of $ \cos(1/z) $ simple? $\endgroup$ – Alexandar Solženjicin Jan 20 '18 at 0:06

If $f$ has a pole at $a$ then the order of the pole is 1 if and only if $(z-a)f(z)$ has a finite limit as $z \to a$. To see that this is indeed the case just apply L'Hopital's Rule.


Taking the expression of $\cos {z}$ as product $$ \cos \left( {1/z} \right) = \;\;\prod\limits_{k = 1}^\infty {\left( {1 - {1 \over {\left( {\pi \,z\left( {k - 1/2} \right)} \right)^{\,2} }}} \right)} $$

then it is clear that $1/\cos{1/z}$ has simple poles in $$ z = \pm {1 \over {\pi \left( {k - 1/2} \right)}}\quad \left| {\;k \in \mathbb N} \right. $$


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