Laurent series for $\frac{1}{z(z+3)(z-1)^2}$ Find the Laurent Series for 
$$\frac{1}{z(z+3)(z-1)^2}$$ in $1 < |z-1| < 4$
So I did the partial fraction decomposition which yields:
$$\frac{1}{4(-1+z)^2} - \frac{5}{16(-1+z)} + \frac{1}{3z} - \frac{1}{48(3+z)}$$
Can anyone help me finish this problem?

You are on the right track.  Partial fraction expansion reveals
$$\frac{1}{z(z+3)(z-1)^2}=\frac{1/3}{z}-\frac{1/48}{z+3}-\frac{5/16}{z-1}+\frac{1/4}{(z-1)^2}\tag1$$
The last two terms on the right-hand side of $(1)$ constitute part of the Laurent expansion in the annulus $1<|z-1|<4$.
We now expand the first term, $\frac{1/3}{z}$, for $|z-1|>1$ as
$$\begin{align}
\frac{1/3}{z}&=\frac{1/3}{z-1}\left(\frac{1}{1+\frac{1}{z-1}}\right)\\\\
&=\frac13\sum_{n=0}^\infty (-1)^n (z-1)^{-(n+1)}\tag2
\end{align}$$
Similarly, we expand the term $\frac{1/48}{z+3}$ for $|z-1|<4$ as 
$$\begin{align}
\frac{1/48}{z+3}&=\frac{1/48}{4+(z-1)}\\\\
&=\frac1{192}\sum_{n=0}^\infty (-1)^n\left(\frac{z-1}{4}\right)^n\tag3
\end{align}$$ 
Now, assemble the expansion using $(2)$ and $(3)$ in $(1)$.
