I find myself struggling a bit with Laurent series, and I'm currently stuck on this exercise below.
I want to find the coefficients $c_{-1}, c_0, c_1, c_2, c_3$ of the Laurent series of $\frac{1}{\sin(z)}$ valid in $A(0,0,\pi)$.
My solution so far:
- Investigate area, and find what singularities we have.
$A(0,0,\pi)$ gives that we have $-\pi,0,\pi$ as singularities.
- Write as a Laurent series.
$$\sin(z)= z-\frac{z^3}{6}+\frac{z^5}{120}+O(z^7) \\\Rightarrow \frac{1}{\sin(z)} = \frac{1}{z-\frac{z^3}{6}+\frac{z^5}{120}+O(z^7)} \\ = \frac{1}{z} \frac{1}{1-\frac{z^2}{6}+\frac{z^4}{120}+O(z^6)} \\ = \frac{1}{z} \frac{1}{1-(\frac{z^2}{6}-\frac{z^4}{120}+O(z^6)}$$
$$ \frac{1}{z}\sum_0^{\infty} \Big(\frac{z^2}{6}-\frac{z^4}{120}+O(z^6)\Big)^k$$
I would guesss that the next step is to somehow write out the sum again to find each coefficients, but I'm kind of lost at this point.
Would anyone like to give me a hint?