The question is: Find the inverse z-transform of $E(z)=\frac{z}{(z-2)(z-1)^2}$
so far I have: using partial fractions with repeating factors... $$\frac{z}{(z-2)(z-1)^2}=\frac{Az}{(z-2)}+\frac{Bz}{(z-1)}+\frac{Cz}{(z-1)^2}\\ z=Az(z-1)^2+Bz(z-2)(z-1)^2+Cz(z-2)$$
z=1: $$1=A1(1-1)^2+B(1(1-2)(1-1)^2+C1(1-2)^2\\ 1=0+0+C(-1)^2\\ 1=-C\\ C=-1$$
setting z=2 results in: $$A=1$$
How do I go about solving B?
$$E(z)=\frac{z}{z-2}+\frac{Bz}{z-1}-\frac{z}{(z-2)^2}$$ taking the inverse z-transform then... $$e(k)=2^k+???+\frac{1}{2}k2^k$$
Please tell me this is remotely correct. If not, please show me where I am going wrong.
Your assumption that $\frac{z}{(z-2)(z-1)^2}=\frac{Az}{(z-2)}+\frac{Bz}{(z-1)}+\frac{Cz}{(z-1)^2}$ is wrong. It should be: $$\frac{z}{(z-2)(z-1)^2}=\frac{A}{(z-2)}+\frac{B}{(z-1)}+\frac{C}{(z-1)^2}$$
See here for examples why.
Now, multiply by the denominator on the left hand side to get $$z=A(z-1)^2+B(z-2)(z-1)+C(z-2) \implies\\ (A+B)z^2-(1+2A+3B+C)z+A+2B-2C=0$$
For this equation to generally hold, the coefficients of each of the powers of $z$ must be equal on both sides (the coefficients on the right hand side is of course zero). This leads to the following equation system.
$$\begin{cases} A+B=0\\1+2A+3B-C=0\\A+2B-2C=0\\ \end{cases} $$ Now solve it, and plug in the resulting $A$, $B$ and $C$ above and apply the inverse z-transform on the terms.