complex numbers - Calculate Laurent series for $1/ sin(z)$

How can calculate Laurent series for

$$f(z)=1/ \sin(z)$$ ??

I searched for it and found only the final result, is there a simple way to explain it ?

Answer

Using the series for $\sin(z)$ and the formula for products of power series, we can get
$$\begin{align} \frac1{\sin(z)} &=\frac1z\frac{z}{\sin(z)}\\ &=\frac1z\left(1-\frac{z^2}{3!}+\frac{z^4}{5!}-\frac{z^6}{7!}+\cdots\right)^{-1}\\ &=\frac1z\left(1+\frac{z^2}{6}+\frac{7z^4}{360}+\frac{31z^6}{15120}+\cdots\right)\\ &=\frac1z+\frac{z}{6}+\frac{7z^3}{360}+\frac{31z^5}{15120}+\cdots \end{align}$$

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Using the formula for products of power series

As given in the Wikipedia article linked above,

$$\left(\sum_{k=0}^\infty a_kz^k\right)\left(\sum_{k=0}^\infty b_kz^k\right) =\sum_{k=0}^\infty c_kz^k\tag{1}$$
where
$$c_k=\sum_{j=0}^ka_jb_{k-j}\tag{2}$$
Set
$$c_k=\left\{\begin{array}{} 1&\text{for }k=0\\ 0&\text{otherwise} \end{array}\right.\tag{3}$$
and
$$a_k=\left\{\begin{array}{} \frac{(-1)^j}{(2j+1)!}&\text{for }k=2j\\ 0&\text{for }k=2j+1 \end{array}\right.\tag{4}$$
Using $(2)$, $(3)$, and $(4)$, we can iteratively compute $b_k$.

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For example, to compute the coefficient of $z^8$:
$$\begin{align} c_8=0 &=b_8-\frac16b_6+\frac1{120}b_4-\frac1{5040}b_2+\frac1{362880}b_0\\ &=b_8-\frac16\frac{31}{15120}+\frac1{120}\frac7{360}-\frac1{5040}\frac16+\frac1{362880}1\\ &=b_8-\frac{127}{604800} \end{align}$$
Thus, $b_8=\dfrac{127}{604800}$.