# ROC of Z transform Doesnt include a pole on the boundary?

I cannot figure out what is going on here. I have an example problem in my book that says the ROC of a certain function is $$0.5 < |z| < \infty$$ The function's denominator is $$1 - z^{-1} + 0.75z^{-2} -0.25z^{-3} + 0.0625z^{-4}$$ By properties of the ROC, there should be a pole on the boundary of the ROC. Since the example given was a right sided signal, the ROC will be the area outside of a circle boundary. So, the signal should have a pole on the boundary of that circle, which has a radius of 0.5 correct?

But when I plug in 0.5 to the denominator equation, I get 1, not zero, which makes 0.5 not a pole in this case. I have tried graphs, and placing different parenthesis, and calculated it multiple times, but the denominator equation above never reaches zero. According to ROC properties, it has to though!

Where am I thinking wrong here?

Thanks,

-Dom

The ROC $$0.5<|z|<\infty$$ does not imply that there's a pole at $$z=0.5$$. What it does say is that there's at least one pole satisfying $$|z|=0.5$$ (and no other pole with a radius larger than $$0.5$$), and this is also the case for the given denominator polynomial. The roots of the polynomial are
$$z_{1,2}=0.5\, e^{j\pi/3}\tag{1}$$
$$z_{3,4}=0.5\, e^{-j\pi/3}\tag{1}$$
I.e., there are two double poles, and all four poles lie on a circle with radius $$0.5$$.