Laurent Series for specified domain 
Find the first three nonzero terms of the Laurent series for $\frac{z}{(z-1)(z-3)^4}$ in the following domains:
  (a) $0<|z-3| <2$ 
(b) $ 2 < | z-3|$

I'm a little confused on how to do this. My attempt is below:
Part (a) : $\frac{1}{(z-3)^4}$$[1 + \frac{1}{z-1}] = \frac{1}{(z-3)^4}[1 + \frac{1}{2}(\frac{1}{1 + \frac{z-3}{2}})]$ $= $ $\frac{1}{(z-3)^4} + \frac{1}{2(z-3)^4}\sum_{k=0}\frac{(-1)^k(z-3)^k}{2^k}$
This is as far as I've gotten. 
So to find the first three nonzero terms, I just subsitute 0, 1, 2 in for $k$ since I created the series around the domain for (a)?
The first three nonzero terms I get in part (a) is $\frac{3}{2(z-3)^4} $ for $ k = 0,  \frac{-z +7}{4(z-3)^4} $ for $ k = 1, $ and $ \frac{8 + (z-3)^2}{8(z-3)^4} $ for $ k = 2$.
Is this correct? I know they're nonzero because of the domain condition.
 A: Here’s how I do it, after noticing that both annuli are centered at $3$:
For part (a), you want to expand in powers of $z-3$, so you might as well call $z-3=w$, $z=w+3$, and transform your function to an expression in $w$:
$$
\frac{w+3}{w^4(w+2)}=\frac{w^{-4}}2\frac{w+3}{1+\frac w2}\,,
$$
in which the first fraction poses no problems, and the second expands out as $(w+3)(1-\frac w2+\frac{w^2}4-\frac{w^3}8+\cdots)$. You can finish this one off, I’m sure. Note that the geometric series in the last pair of parentheses has common ratio $-w/2$, thus is convergent for $|w/2|<1$, in other words $|w|<2$, exactly right.
For part (b), your annulus is unbounded, so you want powers of $w^{-1}$, and you might as well call this by a different letter to make everything clearer, let’s call $1/w=t$, so the expression becomes
$$
t^4\frac{\frac1t+3}{\frac1t+2}=t^4\frac{1+3t}{1+2t}=t^4(1+3t)(1-2t+4t^2-8t^3+\cdots)\,,
$$
and now this geometric series has common ratio $-2t$, so is convergent only for $|2t|<1$, thus for $|2/w|<1$, in other words $|w|>2$, again exactly right.
