If a rational function is real on the unit circle, what does that say about its roots and poles?

While doing a bit of self study, I ran across a situation whose wording confused me.

Suppose $R(z)$ is some rational function which is real on the circle $|z|=1$ in the complex plane. The question asks, how are the zeros and poles situated?

I don't quite understand this, what does it mean by how they're "situated"? Is there some trick I'm supposed to use here? Does a nice scenario pop out, like they're reflections across the origin or the axes from each other or something similar?

From the comments and help, I think I've made a little progress.

If $R(c)=0$, then $\overline{R(1/\bar{c})}=0$, and thus $R(1/\bar{c})=0$ by conjugating again. So switching the roles shows $c$ is a root iff $1/\bar{c}$ is a root, and $c$ is a pole iff $1/\bar{c}$ is a pole? And I think the geometric description to this situation is that inversion in the unit circle preserves poles and roots.

Also, the roots and poles come in pairs except when $c=1/\bar{c}$, that is, when $|c|=1$. So the only thing I can think of is that the roots come in pairs off the unit circle, and the poles come in pairs off the unit circle, plus some possible unpaired roots and poles on the unit circle. To those more experienced, does this seem like the intended answer to how the roots and poles are situated?

Many thanks,

Solution 1:

Yes, there should be some symmetries of the zeros and poles. Notice that $1/\bar z = z$ on the unit circle; therefore $R(1/\bar z) = R(z)$ automatically. Since $R$ is real on the unit circle, we also have $\overline{R(1/\bar z)} = R(z)$. But both sides are rational functions....