What is the inverse z transform of 1/(z-1)^2? I'd like to know how to calculate the inverse z transform of $\frac{1}{(z-1)^2}$ and the general case $\frac{1}{(z-a)^2}$
 A: Hint: find the Maclauren series expansion of $\frac{1}{(z-a)^2}$ by differentiating a power series term-by-term.
Solution: Recall that for a sequence $x[n], n\in\mathbb{Z}$, the $\mathcal{Z}$ transform is defined as 
$$
\mathcal{Z}\{x[n]\}(z)=\sum_{n\in\Bbb{Z}}x[n]z^{-n}
$$If we can find a power series expansion for $(x-a)^{-2}$, the coefficients of our power series will be the sequence $x[n]$.  This will be easier than using definition of the inverse transform.
Now, for any $a\neq 0$, we have 
$$
\frac{1}{z-a}=-\frac{1}{a}\frac{1}{1-z/a}=-\frac{1}{a}\sum_{n=0}^\infty\left(\frac{z}{a}\right)^n
$$Hence, inside the radius of convergence, we have 
$$
(z-a)^{-2}=-\frac{d}{dz}(z-a)^{-1}=\sum_{n=1}^\infty\frac{n}{a^{n+1}}z^{n-1}
$$ 
We now have a power series, so the only thing that remains is to define the appropriate $x[n]$ such that $\mathcal{Z}\{x[n]\}=(z-a)^{-2}$.  Note that we have only nonnegative powers of $z$, so if we define 
$$
x[n]=\left\{\begin{array}{cc}
\frac{-n+1}{a^{-n+2}} & n\leq 0\\
0 &n>0
\end{array}\right.
$$ we have $\mathcal{Z}\{x[n]\}=(z-a)^{-2}$, as you can check. 
A: A simple way is to recall the convolution property: if $x[n] \leftrightarrow X(z)$, then 
$x[n] \star y[n] \leftrightarrow X(z) Y(z) $ and, in particular $x[n] \star x[n] \leftrightarrow [X(z)]^2$
Then, we only need to get the anti-transform of $\frac{1}{z-a}= \frac{ z^{-1}}{1-a z^{-1}}$  (piece of cake) and convolve it with itself. 
BTW, you need to specify the ROC (region of convergence), or if you want a "causal" sequence - without this, there can be more than one solution.
