Find the Laurent Series of $\frac{z^2}{(z-1)(z^2+4)}$ I have attempted to split this into partial fractions of the form $\frac{A}{(z-1)} + \frac{B}{(z-2i)} + \frac{C}{(z+2i)}$ as well as by $\frac{A}{(z-1)} + \frac{Bz+C}{(z^2+4)}$, but things get very complex (ha!) with i's and z's showing up in the numerator. How should I proceed?
I am told to find the series about 0 and 1, taking into account the different values of the radii.

Since you are told to find the series about $0$ and $1$, the first one does not make much trouble using long division.
For the second one, define first $z=x+1$ which makes 
$$f=\frac{z^2}{(z-1)(z^2+4)}=\frac{(x+1)^2}{x \left(x^2+2 x+5\right)}=\frac{x^2+2x+1}{x \left(x^2+2 x+5\right)}$$ Perform the long division for $$x \times f=\frac{x^2+2x+1}{x^2+2 x+5}$$ When done, divide the result by $x$ and replace $x$ by $(z-1)$
