Laurent series for this function

[Carl140]

Homework Statement

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

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?

 

[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.