Finding residue of complex function, the result is different when using Laurent series and residue theorem. Find the residue of
$$f(z)=\dfrac{z^3+2z+1}{(z-1)(z+3)}$$
on simple pole $z=1$.
If I using residue theorem, I have
\begin{align}
\underset{z=1} {\operatorname{Res}} f(z) =
\lim\limits_{z\to 1} (z-1)\dfrac{z^3+2z+1}{(z-1)(z+3)}=\dfrac{1+2+1}{4}=1.
\end{align}
If I using Laurent series method, I have
\begin{align}
f(z)&=\dfrac{z^3+2z+1}{(z-1)(z+3)}\\
&=(z-2)+\dfrac{9z-5}{(z-1)(z+3)}\\
&= -1+(z-1)+\dfrac{9(z+3)-32}{(z-1)(z+3)}\\
&= -1+(z-1)+\dfrac{9}{(z-1)}-\dfrac{32}{z-1}\cdot\dfrac{1}{z+3}\\
&= -1+(z-1)+\dfrac{9}{(z-1)}-\dfrac{32}{z-1}\cdot\dfrac{1}{z-1+4}\\
&= -1+(z-1)+\dfrac{9}{(z-1)}-\dfrac{32}{(z-1)^2}\cdot\dfrac{1}{1+\dfrac{4}{z-1}}\\
&= -1+(z-1)+\dfrac{9}{(z-1)}-\dfrac{32}{(z-1)^2}\sum\limits_{n=0}^{\infty} (-1)^n \left(\dfrac{4}{z-1}\right)^n\\
&= -1+(z-1)+\dfrac{9}{(z-1)}-\dfrac{32}{(z-1)^2}\sum\limits_{n=0}^{\infty} (-4)^n \left(z-1\right)^{-n}\\
&= -1+(z-1)+\dfrac{9}{(z-1)}+\sum\limits_{n=0}^{\infty} (-32)(-4)^n \left(z-1\right)^{-n-2}.
\end{align}
Now I have coefficient of $(z-1)^{-1}$ is $9$, so we can conclude
\begin{align}
\underset{z=1} {\operatorname{Res}} f(z) =9.
\end{align}
My question
When I using residue theorem and Laurent series method, why the result is distinct? What my mistake in my work?

The "Laurent series method" had an incorrect conclusion going from the fourth to the fifth lines. There was a still an order $1$ term on the far right, it just still had to be pulled out. 
We can calculate the Laurent series another way. From partial fraction decomposition we have that
$$\frac{4}{(z-1)(z+3)} = \frac{1}{z-1}-\frac{1}{z+3}$$
which means we can rewrite the function as
$$f(z) = \frac{z^3+2z+1}{4(z-1)} - \frac{z^3+2z+1}{4(z+3)}$$
This time the term on the far right has no singularity at $z=1$ so it will not contribute to the residue. We will only focus on the term on the left.
Next, shift the variables in the numerator to the desired center:
$$\frac{z^3+2z+1}{4(z-1)} = \frac{(z-1)^3+3(z-1)^2+5(z-1)+4}{4(z-1)}$$
$$ = \frac{1}{4}(z-1)^2+\frac{3}{4}(z-1) + \frac{5}{4} + \frac{1}{z-1}$$
which has a residue of $1$.
