Laurent series of $\frac{1}{1-\cos{3z}}$

Question

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 $0 < |z| < R$.

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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

Answer 1

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 $1-\cos(3z)=0$.

Answer 2

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}