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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,

1 Answers 1

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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....

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    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?2012-01-24
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    @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](http://en.wikipedia.org/wiki/Inversive_geometry#Circle_inversion).2012-01-24
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    @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?2012-01-24
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    @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.2012-01-24
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    @JonasMeyer Whoops, I totally messed that up. So if $R(c)=0$, then $\overline{R(1/\bar{c})}=0$, and thus $R(\bar{c}^{-1})=0$ by conjugating again. So switching the roles shows $c$ is a root iff $\bar{c}^{-1}$ is a root, and $c$ is a pole iff $\bar{c}^{-1}$ is a pole? And the geometric description is that inversion in the unit circle preserves poles and roots?2012-01-24
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    Since $\mathbb{C}$ is algebraically closed, $R(z)$ can be expressed as a constant times a finite product of nonzero integral powers of monic linear terms: $R(z)=a\prod(z-z_k)^{n_k}$. Each $z_k$ is a root or a pole, depending on the sign of $n_k$. Can you conclude something about the set $\{z_k\}$ using a pairing argument about the zeros and poles of R? Perhaps you will want to write $z=rw$ for $r\in\mathbb{R}$ and $w$ on the unit disc ($|w|=1$).2012-01-24
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    @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?2012-01-24
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    @majora: yes, that's what I had in mind, but I haven't thought about it deeply.2012-01-24
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    @bgins Thanks, I think I'm getting a better idea of what I'm looking for.2012-01-24
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    If we are concluding that $c$ and $1/\overline{c}$ are simultaneously roots or poles, where did we need to use that $R(z)$ is real on the unit circle? It seems that this is true without this extra condition?2018-07-10
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    Well, no: $R(z)=z-1/2$ is a rational function, but it has a zero at $1/2$ yet not at $2$. The step where the real-valuedness is used is explicit in my answer. It is helpful to remember that if $f(z)$ is a meromorphic function, then $f(\bar z)$ is not necessarily meromorphic, but $\overline{f(\bar z)}$ is (check the Cauchy–Riemann equations).2018-07-11