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Regarding complex functions (in complex variables), I was wondering why the function $g(z)= \cos(z)$ has a singularity at $z = \infty$ but $f(z)= \dfrac{z}{\cos(z)}$ does not.

I am a bit confused about the concept of singularities at $\infty$.

Thank you.

Michael Hardy
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Merry
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    Both of those functions have essential singularities at $z=\infty$. A function $f(z)$ has a singularity at $z=\infty$ if $f(1/w)$ has a singularity at $w=0$. – Antonio Vargas May 28 '13 at 04:05
  • ah it does? One of the examples I was looking at was $g(z) = \frac{(z^2-1)(z-3)^3}{sin(z)}$ and it said that it only had singularities at the points $z$ where $sin(z)$ was zero. I thought the function above would also not have a singularity at infinity, but I am confused on why for g, this is the case then. – Merry May 28 '13 at 07:32
  • In most situations, complex analysis deals with $\mathbb{C}$, not the Riemann sphere, so most of the time you don't care about what happens at infinity. – mrf May 28 '13 at 08:06
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    @AntonioVargas: This seems to be a common confusion, but, no, $f(z)$ does not have an essential singularity at $\infty$. By definition, an essential singularity must be isolated and this function has poles at the points $z_n=n\pi+\pi/2$, with $z_n\to\infty$. – Ted Shifrin May 28 '13 at 14:19
  • @TedShifrin you're right, I forgot about that. Perhaps there should be another word for non-isolated singularities? Terrible singularities? – Antonio Vargas May 28 '13 at 14:25
  • @AntonioVargas: Indeed :) They certainly wreak havoc as they want! – Ted Shifrin May 28 '13 at 14:26

2 Answers2

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So, to sum up after all these comments:

$g(z)$ indeed has an essential singularity at $z=\infty$ because the function $h(z)=g\big(\frac1z\big)$ has an essential singularity at $z=0$: $$h(z) = \cos\left(\frac1z\right) = 1- \frac1{2!}\left(\frac1z\right)^2 + \frac1{4!}\left(\frac1z\right)^4+ \dots $$ has infinitely many terms in its Laurent expansion at $0$.

On the other hand, $f(z)$, all appearances to the contrary, does not have an essential singularity at $\infty$: $f(z)$ has poles at $z=z_n=n\pi + \pi/2$ and $z_n\to\infty$, and so $\infty$ is not an isolated singular point.

Ted Shifrin
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For $\,\frac1z\,$ "close enough to zero" ( say, $\;\frac1{|z|}<\sqrt[3]6\;$ ) , we have

$$f\left(\frac1z\right)=\frac1{z\cos\frac1z}=\frac1z\frac1{\left(1-\frac1{2z^2}+\frac1{24z^4}-\ldots\right)}=$$

$$=\frac1z\left(1+\frac1{2z^2}+\frac1{4z^4}+\ldots\right)=\frac1z+\frac1{2z^3}+\ldots$$

which shows $\,f(z)\;$ has a singularity at $\,z=\infty\;$

DonAntonio
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  • I think we have the same problem here as in the other question (http://math.stackexchange.com/questions/404631/finding-the-singularity-type-at-z-0-of-frac1-cos-frac1z/404668?noredirect=1#comment865900_404668). What the OP stated is correct: $\infty$ is not an isolated singularity of $\dfrac1{\cos z}$ because there is a sequence of singular points converging to it. I'm not sure what's going on with your Laurent series here. – Ted Shifrin May 28 '13 at 13:02
  • In the series for $\cos(1/z)$ you forgot to replace $z$ by $1/z$. Also $\cos$ is even so its series should only have even powers of $z$. You will end up with infinitely-many negative powers of $z$. – Antonio Vargas May 28 '13 at 13:55
  • Thanks @AntonioVargas, corrected – DonAntonio May 28 '13 at 15:35