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.
On
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 $(z-1)$
Hint: use the limits $\lim_{z\to 1} (z-1)f(z) = \frac{1^2}{1^2+4}$ and similarly $(z-2i)f(z)$ for $z\to 2i$ and $(z+2i)f(z)$ for $z\to -2i$. Also note that $C = \overline B$ formally.