Question about poles and zeros in AR filter

For AR HP filter

1. zeros, to the right of the imaginary axis
2. poles outside the unit-circle
3. zeros on the real axis
4. poles, to the left of the imaginary axis

Apparently, the right answer is (4). Why?

We did learn in class some properties of the zeros and poles, and I probably need to relate it to the fact that it's a AR HP filter.

Thanks!

EDIT:
I know that AR is an all-poles filter so it must be (1) or (4). What is the meaning of the position of the pole with relation to the imaginary axis?

EDIT 2:
Is it because the large $\omega$'s (frequencies as angles of the unit circle) are at the left side of the imaginary axis and so we get an HP filter?

• In the continuous case, $s$ such that $|s|>\sigma$ for some $\sigma\in\mathbb{R}$.
• In the discrete case, $z$ such that $|z|>r$ for some $r\in\mathbb{R}$.
Regarding the poles, it depends on the coefficient of the process, $a$. If $|a|<1$, the filter will be stable and all the poles will be on the left half plane of the $s$-plane, or inside the unit circle in the $z$-domain. If $|a|>1$, the contrary will happen.