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}
$$

---

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 $b_k$.

---

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, $b_8=\dfrac{127}{604800}$.