Laurent series of $\log(z)\sin(1/(z-1))$ I would like to calculate the Laurent series of $$\log(z)\sin \left(\frac{1}{z-1} \right)$$ I have developed  separately $\log z$ et the sinus and tried to multiplied them terms by terms but it is complicated. Is there another simplier way?

Hints and a bit of example
Use the substitution
$$z - 1 = u$$ 
so that
$$\sin\left(\frac{1}{z-1}\right) = \sin\left(\frac{1}{u}\right) = \frac{1}{u} - \frac{1}{3!u^3} + \cdot$$
and
$$\log(z) = \log(u+1) = u - \frac{u^2}{2} + \frac{u^3}{3} - \cdot $$
In this way you get simply
$$\left(\frac{1}{u} - \frac{1}{3!u^3}\right)\cdot \left(u - \frac{u^2}{2} + \frac{u^3}{3}\right) = -\frac{1}{6u^2} + \frac{1}{12u} + \frac{17}{18} - \frac{u}{2} + \frac{u^2}{3}$$
