I have to find the Laurent-Series of the following function:
$$f(z) = \frac{\sin({\frac1{iz^3}})}{z^2} $$ in $$z_0=0$$
My approach:
Using: $$ \sin(z) = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} {z^{2n+1}} $$
$$f(z) = \frac{\sin({\frac1{iz^3}})}{z^2} = \frac{\sin({\frac{-i}{z^3}})}{z^2} = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} {(-iz)^{-6n-5}} $$
I can't visualize the residue (coeffincent ${c_{-1}}$) and I have no idea how to find the valid region.
Thank you.
The Laurent series is sum from n=0 To infinity of (-1)^n /(2n+1)! * 1/i^2n+1 * 1/z^6n+5. The residues is when 6n+5=1 Which is never for integer n, so the residue is 0.