I have trouble finding a (probably) pretty easy inverse of a $z$ transform.
$$H(z) = \frac{z-0.5}{z+0.5}$$
I used the polynomial division on it to get a proper fraction and got $$H(z) = 1 - \frac{1}{z+0.5}$$ took the $z$ out $$H(z) = 1- z^{-1} \cdot \frac{1}{1+0.5z^{-1}}$$
Now I have trouble finding the inverse of $\frac{1}{1+az^{-1}}$. I have looked it up in my transform tables, but only find the transform for the case $\frac{1}{1-az^{-1}}$ which is $a^{n}$.
Matlab says it's $(-1)^{n}$, but I'm quite new to matlab and don't trust anything I put in. Does anyone know the transformation?
If you express $\frac{1}{1+az^{-1}}$ as $\frac{1}{1-(-az^{-1})}$, then the fraction expands to $$\frac{1}{1-(-az^{-1})}=\sum_{n=0}^{\infty}(-a)^n(z^{-n})$$ so the inverse transform will be $(-a)^n$.