Given $f(z) = \dfrac{1}{\sin(z)}$
a) Give singularities
b) Determine coefficients $a_{-1}$ and $a_{-3}$ of the Laurent series
So I thought:
a) $n \pi$, where $n$ is an integer
b) $\dfrac{1}{\sin(z)}$ can be rewritten as $\dfrac{1}{z(1-1/6z^2+...)}$ so $a_1$ is $\dfrac{1}{6}$ but how can I find $a_{-1}$ and $a_{-3}$ ?
Thanks !
$$\sin z\frac1{\sin z}=(z-\frac{z^3}6+\frac{z^5}{120}-...)(\frac1z+\frac16z+a_3z^3+...)=1$$
What is the coefficient of $z^4$?