I need to calculate principal part of the Laurent series expansion of $f$ at $z_0=0$ with $$ f(z)=\frac{\sin(z^3)}{(1-\cos z)^3} $$
I can see that $f$ has a pole of order 3 at $z_0=0$ , and also that $f$ is an odd function, therefore the expansion will be $$ a_{-3}z^{-3}+a_{-1}z^{-1} $$ but after this I am stuck. I tried using some trigonomteric identities, but they seem to make things more complicated.
I calculated $a_{-1}$ using the residue formula and it seems to be 2 , but I think there has to be a simpler way to deal with this.
Any help or advice towards the soluution would be welcome.
Hoping that the Taylor series could help you to some extent, let us start with $$\sin(y)=y-\frac{y^3}{6}+O\left(y^4\right)$$ So $$\sin(z^3)=z^3-\frac{z^9}{6}+O\left(z^{10}\right)$$ Now $$\cos(z)=1-\frac{z^2}{2}+\frac{z^4}{24}+O\left(z^5\right)$$ $$1-\cos(z)=\frac{z^2}{2}-\frac{z^4}{24}+O\left(z^5\right)$$ So $$\frac{\sin(z^3)}{(1-\cos z)^3}=\frac{8}{z^3}+\frac{2}{z}+\frac{4 z}{15}+O\left(z^3\right)$$