# Laurent series for this function

1. Jan 9, 2009

### Carl140

1. The problem statement, all variables and given/known data

Find the Laurent series of the function f(z) = Sin(1/(z^2-z)) in the region 0<|z|<infinity.

3. The attempt at a solution

Now sin(z) = [e^(iz) - e^(-iz)]/(2i)

Shall we replace z by 1/(z^2-z) to obtain the Laurent series for f(z)?
I tried this but it gets messy. Is there a clever method? or any other approach?

2. Jan 9, 2009

### daschaich

I would try using the series expansion
$$\sin(z) = z - \frac{z^3}{3!} + \frac{z^5}{5!} - \cdots$$
without going through the exponentials. That should be cleaner, but I haven't checked and I don't know if it's the best way to go.