How to apply cauchy's residue theorem to rational function? I have the following function:
$$f(z)=\frac{(z-1)^2}{z*(z+1)^3)}$$
And I need to find its residue.  
The formula that I know for this is the following:
The residue $a_{-1}$ for a pole of order "n" is:
$$a_{-1} =\frac{1}{(n-1)!} \lim_{z \rightarrow z_0} \frac{\partial^{n-1}(z-z_0)^{n}(f(z))}{\partial z^{n-1}}$$
Now, when I apply this to the function $f(z)$ above, I'm getting the following:
$$a_{-1} =\frac{1}{(3-1)!} \lim_{z \rightarrow -1} \frac{\partial^{3-1}(z-z_0)^{3}(f(z))}{\partial z^{3-1}}$$
$$a_{-1} =\frac{1}{2} \lim_{z \rightarrow -1} 2z - 2 = -2$$ 
However, I'm told that the answer is $-4$.  Where did I go wrong in my solution?
Thanks.

I get $-1$ for the residue.  $$(z+1)^3f(z)= \frac{(z-1)^2}{z}=z-2+\frac1z$$  The second derivative is $2z^{-3}$ and $\lim_{z\to-1}\frac12\cdot2z^{-3}=-1$.
Alternatively, we want the coefficient of the $\frac{1}{z+1}$ term in the Laurent series fo $f(z)$.  We have $$\frac{(z-1)^2}{z} = z-2+\frac1z = (z+1)-3-\frac{1}{1-(z+1)}$$  Dividing by $(z+1)^3$, we see that we want the coefficient of $(z+1)^2$ in the Taylor series for $-\frac{1}{1-(z+1)}$, which is $-1$.
