Laurent series of $\log(z)\sin(1/(z-1))$

Question

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?

Answer 1

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

$$\left(\frac{1}{u} - \frac{1}{3!u^3}\right)\cdot \left(u - \frac{u^2}{2} + \frac{u^3}{3}\right) = -\frac{1}{6u^2} + \frac{1}{12u} + \frac{17}{18} - \frac{u}{2} + \frac{u^2}{3}$$

Answer 2

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

$$a_{-1} = 2 \sum_{k=1}^{\infty} \frac1{(4 k)!} = \left (\cosh{1} + \cos{1} \right ) - 2 = 2 \left (\sinh^2{\frac12} - \sin^2{\frac12} \right )$$