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?
 A: Assuming we expand about $\zeta=z-1$, then
$$\log{(1+\zeta)} = \sum_{k=1}^{\infty} \frac{(-1)^k \zeta^k}{k} $$
$$\sin{\frac1{\zeta}} =  \sum_{n=0}^{\infty} \frac{(-1)^n}{(2 n+1)!\,\zeta^{2 n+1}} $$
There really is no substitute for multiplying the series out.  The best you can do is write
$$\log{(1+\zeta)} \sin{\frac1{\zeta}} = \sum_{m=-\infty}^{\infty} a_m \zeta^m $$
and compute the coefficients $a_m$ as needed.  As an example, let's compute the residue, or the coefficient of $\zeta^{-1}$.  By multiplying the terms out, you can show that the residue is
$$a_{-1} = 2 \sum_{k=1}^{\infty} \frac1{(4 k)!}  =  \left (\cosh{1} + \cos{1} \right ) - 2 = 2 \left (\sinh^2{\frac12} - \sin^2{\frac12} \right )$$
A: 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
$$\left(\frac{1}{u} - \frac{1}{3!u^3}\right)\cdot \left(u - \frac{u^2}{2} + \frac{u^3}{3}\right) = -\frac{1}{6u^2} + \frac{1}{12u} + \frac{17}{18} - \frac{u}{2} + \frac{u^2}{3}$$
