Find the Laurent series for $f(z) = (z^2 - 4)/(z-1)^2 $ for $z=1$ What I understand is that we have to expand $f(z$) in the positive and negative powers of $(z-1)$.
Hence I tried factorizing the numerator $(z^2-4)=(z+2)(z-2)$ , which can then be written in terms of $(z-1)$ as:
$(z-1-1)(z-1+3)/(z-1)^2$ . however i cannot expand it using the geometric series expansion:
$$1/(1-z) = 1 + z + z^2 + \cdots$$
Help.

Use
$$z^2-4 = (z-1+1)^2 -4 = (z-1)^2+ 2 (z-1)-3$$
so that
$$f(z) = -\frac{3}{(z-1)^2} + \frac{2}{z-1}+1$$
