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Let the Laurent series of the function $f(z)=\frac{1}{1-\cos{3z}}$ be $\sum_{-\infty}^{\infty}{{a}_{k}{z}^{k}}$.

a) Compute ${a}_{-3}$, ${a}_{0}$ and ${a}_{1}$.

b) Find the biggest $R$ so that the above Laurent series converges in the domain $0 < |z| < R$.

This question is quite similar to the one answered in this post, except we have $3z$ instead of $z$. However I never come up with the same conclusion when doing the calculation myself : \begin{equation} \cos{z} = 1-\frac{{z}^{2}}{2!} + \frac{{z}^{4}}{4!} - \frac{{z}^{6}}{6!} + \dots \end{equation}

\begin{align} \Longrightarrow \frac{1}{1-\cos{z}} &= \frac{1}{\frac{{z}^{2}}{2!} - \frac{{z}^{4}}{4!} + \frac{{z}^{6}}{6!} - \dots} \\ &= \frac{2}{{z}^{2}} \frac{1}{1-\frac{2{z}^{2}}{4!} + \frac{2{z}^{4}}{6!} - \dots} \\ &= \frac{2}{{z}^{2}} \frac{1}{1-\left(\frac{2{z}^{2}}{4!} - \frac{2{z}^{4}}{6!} + \dots\right)}\end{align}

and with $\frac1{1-z}=1+z+z^2+\ldots$ for $|z|<1$, I get :

\begin{align} \frac{1}{1-\cos{z}} &= \frac{2}{z^2}\left(1 + \frac{z^2}{12} - \frac{z^4}{360} + \dots\right) \ &= \frac{2}{z^2} + \frac{1}{6} - \frac{z^2}{180} + \dots \end{align}

I started the Laurent series today for an assignment and therefore I am not very proficient with those objects. However I made the calculation many times and I never get what seems to be the right result from Wolfram. This has probably something to do with the convergence of $\frac{2{z}^{2}}{4!} + \frac{2{z}^{4}}{6!} - \dots$ on the disk $\{z \quad | \quad |z| < 2\pi\}$.

I would love some help about this. Thank you very much

José Carlos Santos
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whytheheckarewedoingmath
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  • There are several answers to this question at https://math.stackexchange.com/a/4672268, https://math.stackexchange.com/a/4672891, https://math.stackexchange.com/a/4672242, and https://math.stackexchange.com/a/4672897 – qifeng618 Apr 04 '23 at 22:02

2 Answers2

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Since $f$ is an even function, $a_{-3}=a_1=0$.

You have $1-\cos(3z)=\frac92z^2-\frac{27}8z^4+\cdots$. So$$\frac1{1-\cos(3z)}=\frac1{\frac92z^2-\frac{27}8z^4+\cdots}=\frac{a_{-2}}{z^2}+a_0+a_2z^2+\cdots$$and therefore$$1=\left(\frac92z^2-\frac{27}8z^4+\cdots\right)\left(\frac{a_{-2}}{z^2}+a_0+a_2z^2+\cdots\right),$$from which you can deduce that $a_{-2}=\frac29$.

The biggest $R$ such that the series converges on $D(0,R)\setminus\{0\}$ is $\frac{2\pi}{3}$, since $\pm\frac{2\pi}3$ are the complex numbers $z$ closest to $0$ such that $1-\cos(3z)=0$.

whytheheckarewedoingmath
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José Carlos Santos
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  • Thank you very much, this is very nice indeed. Just two questions : I don't know how do I justify that $$\frac1{1-\cos(3z)}=\frac{a_{-2}}{z^2}+a_0+a_2z^2+\cdots$$, ie the fact that we have only this negative even power, but I assume this has something to do with poles and singularities, is this where I should look ? And is this the same for the fact that since $f$ is even, then ${a}_{2k+1}=0$ ? – whytheheckarewedoingmath Oct 31 '17 at 10:39
  • @8827 It is just the fact that $f$ is even and nothing else. If my answer was helpful, perhaps that you could mark it as the accepted answer. – José Carlos Santos Oct 31 '17 at 10:45
  • Yes, I'm quite stupid this time --'. My apologies, I'm new here. – whytheheckarewedoingmath Oct 31 '17 at 10:49
  • @8827 I was new here a few months ago and it took me some time until I started to do that. – José Carlos Santos Oct 31 '17 at 10:52
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Hint: \begin{align} \frac{1}{1-\cos3z} &= \frac{1}{1-\left(1-\frac{{(3z)}^{2}}{2!} + \frac{{(3z)}^{4}}{4!} - \frac{{(3z)}^{6}}{6!} - \dots\right)} \\ &= \frac{2}{9{z}^{2}} \frac{1}{1-\frac{9{z}^{2}}{12} + \frac{81{z}^{4}}{360} - \dots} \\ &= \frac{2}{9z^2}+\frac{1}{6}+\frac{3z^2}{40}+\frac{3 z^4}{112}+\cdots \end{align}

Nosrati
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