# What happens when the poles of this z-transform function are outside the ROC for a signal?

I am given a z-transform function for a signal $h[n]$. It's $H(z)=\frac{2z^2-0.75z}{(z-0.25)(z-0.5)}$. I am supposed to find $h[n]$ and check the stability of the system for these cases:

a)ROC: $|z|>0.5$

a)ROC: $|z|<0.25$

a)ROC: $0.25<|z|<0.5$

I found $h[n]$ for the general case to be:

$H(z)=0.5z^{-1}\frac{z}{z-0.5}+0.25z^{-1}\frac{z}{z-0.25}$

$h[n]=(0.5)^nu[n-1] + 0.25^nu[n-1]$

The poles of a function are $0.5$ and $0.25$

The first case is when ROC: $|z|>0.5$. That would mean that i have a circle with radius $0.5$ on the graph on which the causal signals are the ones outside the circle because those values of $z$ make the signal (or a function) converge and anti-causal ones inside the circle. Yet the answer in the book is:

a)$h[n]=(0.5)^nu[n] + 0.25^nu[n]$ and the system is stable. Shouldn't it diverge for the $0.25$ part?

Also when b) ROC: $|z|<0.25$ the answer in the book is:

b) $h[n]=-(0.5)^nu[-n-1] + -0.25^nu[-n-1]$ and the system is unstable.

I could understand this as now the inside of a $0.25$ radius circle is causal and since pole 0.25 is on the circle and 0.5 is outside the circle the $u[n-1]$ is transformed to $u[-n-1]$ to make it causal, but why does the factor $0.25$ and $0.5$ change signes? I will also write down the solution for the c) part:

c) $h[n]=-(0.5)^nu[-n-1] + 0.25^nu[n]$

Can anyone explain these solutions as i'm confused about the whole ROC involvement in these solutions particular.