# 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,

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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....
Thanks. Hmm, so $\overline{R(1/\bar{z})}-R(z)=0$ for all $z$ such that $|z|=1$, but then clearing denominators and such would give a polynomial with infinite roots, so actually $\overline{R(1/\bar{z})}=R(z)$ for all $z$ in the plane? But then what does it mean about how the roots and poles are situated? –  majora Jan 24 '12 at 7:56
@majora: Consider the consequences of this identity when $R(c)=0$ or $R(c)=\infty$. Also note that $z\mapsto 1/\overline z$ has a nice geometric description. –  Jonas Meyer Jan 24 '12 at 8:02
@JonasMeyer So writing $c=a+bi$, it follows that $\overline{R(1/\bar{c})}=\frac{a-bi}{a^2+b^2}$. So if $R(c)=0$, does this imply $a=b=0$? So the roots are situated near the origin? And if $R(c)=\infty$, then I don't quite know what to say about $a$ and $b$, except that they should be very large, so the poles are very far away from the origin? –  majora Jan 24 '12 at 8:23
@majora: No. $R(c)=\overline{R(1/\overline c)}$. So if $R(c)=0$, then _______$=0$. This indicates a symmetry of the zeros that can also be described geometrically. Poles have the same symmetry. –  Jonas Meyer Jan 24 '12 at 8:37
@bgins Thanks, but I don't see a pairing argument between poles and roots. I think that $c$ and $1/\bar{c}$ are simultaneous roots or simultaneous poles, so 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. Is this what you mean? –  majora Jan 24 '12 at 16:01