I want to calculate the Laurent series of $f(z)=\dfrac{\sin(z)}{z \cdot \log(z+1)}$ about $z=0$.
I know $\lim_{z \to 0} \dfrac{\sin(z)}{\log(1+z)}=1$ so f has a pole of order $1$ at $z=0$. Then $a_{-1}=1$ and $a_{-k}=0$ for $k\geq2$
¿What should I do next? I was thinking about multipling $x^{n}f(z)$ and taking derivatives so I can get $a_{0}, a_{1}, a_{2}...$ but I'm not sure that's a good aproach. Any suggestion? Thanks.
Write $\frac{\sin z}{z\ln(1+z)}=\sum_{k\ge0}c_kz^{k-1}$ so $\sum_{k,\,l\ge0}c_k\frac{(-1)^{l+1}}{l+1}z^{k+l}=\frac{\sin z}{z}$ has $z^{2m}$ coefficient $\frac{(-1)^m}{(2m+1)!}$. This gives simultaneous equations in the first few coefficients. The Laurent series begins $\tfrac1z+\tfrac12-\tfrac14z+\tfrac{z^3}{240}+\tfrac{23z^5}{1440}$. I'm not sure if a closed form for the $z^{2n-1}$ coefficient with $n\ge0$ is obtainable.