# 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.