Expand the function $f(z)=\frac{1}{(z-a)(z-b)}$ where $0 < |a| < |b|$ in a Laurent series in different annuli I have to expand the function $f(z) = \frac{1}{(z-a)(z-b)}$ where $a, b \in \mathbb{C}$, $0 < |a| < |b|$ in the following annuli:
(a) $0<|z|<|a|$
(b) $|a|<|z|<|b|$
(c) $|b|<|z|$
I made bonafide attempts at $(b)$ and $(c)$, which I'll share below; however, I was not sure what to do about the $0<|z|$ part in (a). I'm kind of teaching these to myself, and the only examples I've seen with a $0$ in them have involved something like $0< |z+2| <|a|$, or something like that where you'd have to make a substitution $w$... Please let me know how to take care of examples like (a).
Also, here are my attempts at parts (b) and (c). Please let me know if they're correct, and if not, let me know what I need to do in order to fix them:

(b) $\mathbf{|a|<|z|<|b|}$. Using partial fractions, I wrote $\frac{1}{(z-a)(z-b)} = \frac{1}{a-b}\left( \frac{1}{z-a}\right) - \frac{1}{a-b}\left( \frac{1}{z-b}\right)$.  
Now, if $|z|> |a|$, then $\left\vert \frac{a}{z}\right\vert< 1$, so we write $\frac{1}{(a-b)(z-a)} = \frac{1}{(a-b)z(1-a/z)} = \frac{1}{a-b}\cdot \frac{1}{z} \cdot \sum_{n=0}^{\infty} \left( \frac{a}{z}\right)^{n} = \frac{1}{(a-b)}\sum_{n=0}^{\infty}\frac{a^{n}}{z^{n+1}}$.
If $|z|<|b|$, then $\left\vert \frac{z}{b}\right\vert < 1$, so we write $\displaystyle \frac{-1}{(a-b)}\cdot \frac{1}{(z-b)} = \frac{-1}{(a-b)}\cdot \frac{1}{b\left( \frac{z}{b} - 1 \right)} = \frac{1}{b(a-b)}\frac{1}{1-\frac{z}{b}} = \frac{1}{b(a-b)}\sum_{n=0}^{\infty}\left( \frac{z}{b}\right)^{n} = \frac{1}{(a-b)}\sum_{n=0}^{\infty}\frac{z^{n}}{b^{n+1}}$
So, the whole expansion is $\displaystyle \frac{1}{z-b} \left( \sum_{n=0}^{\infty}\frac{a^{n}}{z^{n+1}} + \sum_{n=0}^{\infty}\frac{z^{n}}{b^{n+1}}\right)\\ \displaystyle  = \frac{1}{a-b} \left( \frac{1}{z} + \frac{a}{z^{2}} + \frac{a^{2}}{z^{3}} + \cdots + \frac{1}{b} + \frac{z}{b^{2}}+\frac{z^{2}}{b^{3}}+\cdots\right) \\ \displaystyle =  \frac{1}{z(a-b)} + \frac{a}{z^{2}(a-b)} + \frac{a^{2}}{z^{3}(a-b)} + \cdots + \frac{1}{b(a-b)} + \frac{z}{b^{2}(a-b)}+\frac{z^{2}}{b^{3}(a-b)}+\cdots$
(c) $\mathbf{|b|<|z|}$.  If $|a|<|z|$, we have, as in part (b) $\displaystyle \frac{1}{(a-b)}\frac{1}{(z-a)} = \frac{1}{(a-b)}\sum_{n=0}^{\infty}\frac{a^{n}}{z^{n+1}}$.
If $|b|<|z|$, $\displaystyle \frac{1}{(a-b)}\frac{1}{(z-b)} = \frac{1}{(a-b)} \frac{1}{z \left( 1 - \frac{b}{z}\right)}   \\ \displaystyle = \frac{1}{(a-b)}\frac{1}{z} \sum_{n=0}^{\infty} \left( \frac{b}{z}\right)^{n} \\ \displaystyle = \frac{1}{(a-b)}\frac{1}{z}\left( 1 + \frac{b}{z} + \frac{b^{2}}{z^{2}} + \cdots \right) \\ \displaystyle = \frac{1}{(a-b)} \left( \frac{1}{z} + \frac{b}{z^{2}} + \frac{b^{2}}{z^{3}}+ \cdots\right)$
Then, the required Laurent expansion valid for both $|z|>|a|$ and $|z|>|b|$ is found by subtraction:
$ \displaystyle  \frac{1}{(a-b)} \left( \frac{1}{z} - \frac{1}{z} + \frac{a}{z^{2}} - \frac{b}{z^{2}} + \frac{a^{2}}{z^{3}} - \frac{b^{2}}{z^{3}} + \cdots \right) \\ \displaystyle = \frac{1}{(a-b)} \left( \frac{a-b}{z^{2}} + \frac{a^{2}-b^{2}}{z^{3}} + \frac{a^{3}-b^{3}}{z^{4}} + \cdots \right) \\ \displaystyle = \left( \frac{1}{z^{2}} + \frac{a^{2}-b^{2}}{z^{3}(a-b)} + \frac{a^{3}-b^{3}}{z^{4}(a-b)} + \cdots \right)$

Again, please let me know how to do part (a), as well as if my parts (b) and (c) are ok. Thanks.

Hint:
For $\left|z\right|\lt\left|c\right|$,
$$
\frac1{z-c}=-\frac1c\left(1+\frac zc+\frac{z^2}{c^2}+\frac{z^3}{c^3}+\dots\right)
$$
For $\left|z\right|\gt\left|c\right|$,
$$
\frac1{z-c}=\frac1z\left(1+\frac cz+\frac{c^2}{z^2}+\frac{c^3}{z^3}+\dots\right)
$$
Apply these to
$$
\begin{align}
\frac1{(z-a)(z-b)}
&=\frac1{b-a}\left(\frac1{z-b}-\frac1{z-a}\right)\\
\end{align}
$$
