Inverse Z transform of $\frac {-81}{z^4(z-3)(z-1)}$ The question given like this:

Find the convolution of $3^nu(-n+3)*u(n-2)$

Attempt:
I tried to solve it through z transform method
Let $Y(z)=X1(z)X2(z)$
Now $Z$ transform of $x1(n)=3^nu(-n+3)$ will be $\frac{-81z^{-3}}{z-3}$
Z transform of $u(n-2)=\frac{1}{z(z-1)}$
now $Y(z)=\frac {-81}{z^4(z-3)(z-1)}$ 
I tried to take inverse z transform of this function by residue theorem like this
$Residue at z=3,will be \frac{-3^{n-1}}{2}$.
$Residue at z=1,will be \frac{-81}{2}$
$Residue at z=0,will be -40$
BUT the solution given like this
$$
y(n)=
\begin{cases}
\frac{81}{2},n>=5\\
\frac{3^{n-1}}{2},n<5\\
\end{cases}
$$
Now What mistake i am doing,any other method other than Residue throem will be helpful.
Thanks
 A: I will use the "engineering" approach to find the inverse $Z$ transform by performing a partial fraction expansion and considering each term separately. 
A critical point you have to keep in mind is the region of convergence (ROC) of the $Z$ transform. In particular, since the ROC   is $|z|>1$ and $|z|<3$ for $X_1(z)$ and $X_2(z)$, repsectively, it follows that the ROC for $Y(z)$ must be $1<|z|<3$. 
Performing a partial fraction expansion gives
$$
Y(z)=\frac{81}{2}\frac{z^{-1}}{1-z^{-1}} +\left(- \frac{1}{2}\frac{z^{-1}}{1-3 z^{-1}}\right)+(-40 z^{-1}-39 z^{-2}-36z^{-3}-27 z^{-4}), 1<|z|<3.
$$
Now, by the linearity of the (inverse) Z transform, it follows that $y[n]$ is the sum of three sequences corresponding to the inverse $Z$ transform of each of the three terms of $Y(z)$. These inverse transforms can be easily found using standard tables such as the one provided by wikipedia. 
For example, for the second term, it holds 
$$
\mathcal{Z}^{-1}\left\{ - \frac{1}{2}\frac{z^{-1}}{1-3 z^{-1}}\right\}=\frac{1}{2}3^{n-1} u[-n],
$$
where the transform pair #13 of the table was used.
Remark: Pay attention to the ROC for selecting the right transform formula from the table. #12 has the exact same Z expression as #13, however, its ROC is not compatible with the ROC of $Y(z)$.
It follows that 
$$
y[n] = \frac{81}{2}u[n-1]+\frac{1}{2}3^{n-1} u[-n]-40 \delta[n-1]-39 \delta[n-2]-36\delta[n-3]-27 \delta[n-4].
$$
By inspection, it follows that $y[n]=81/2, n\geq 5$ and $y[n]=3^{n-1}/2,n\leq 0$. Interestingly, evaluating the above expression for $n=1,2,3,4$ shows that $y[n]=3^{n-1}/2$ also for these values of $n$.
A: $$Y(z) = \frac{-81}{z^4(z-3)(z-1)}$$
$$X(z) = z^4Y(z) =\frac{-81}{(z-3)(z-1)}$$
$$X(z) = \frac{-81}{(z-3)(z-1)} = \frac{A}{z-1} + \frac{B}{z-3} $$
$$A = \frac{81}{2}; B=\frac{-81}{2}$$
$$x(n) = \frac{81}{2}[1-3^{n-1}]u(n-1)$$
$$ X(z) = z^4Y(z)$$
$$x(n-4) = y(n)$$
$$y(n) = \frac{81}{2}[1-3^{n-5}]u(n-5)$$
