I'm also self studying the Ahlfors Complex Analysis book.

A question asks:

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 have consulted and understood the argument made here: If a rational function is real on the unit circle, what does that say about its roots and poles?

But, I can't understand why conclusions can be drawn for zeros/poles outside the unit circle: the equation $R(z) = \overline{R(1/\overline{z})}$ should hold only on the unit circle $|z|=1$, so I'm not sure why any conclusion can be made for zeros/poles $\alpha_i$ that are not on the unit circle.

Thanks for your help.


Two rational functions that agree on the unit circle are equal on their common domains, by the identity theorem for holomorphic functions. So the equation in your question holds for all complex numbers. In particular $a$ is a zero of $R$ if and only if $1/\overline{a}$ is a zero of $R$, and similarly $a$ is a pole of $R$ if and only if if $1/\overline{a}$ is a pole of $R$. In both cases the order is preserved.

  • $\begingroup$ Thanks Mark. I actually hadn't encountered the identity theorem yet in the book (wonder why there would be an exercise that used it if it hadn't been covered yet...). But checking the wikipedia article here, your explanation makes sense. $\endgroup$ – anon_student Mar 30 '13 at 20:09
  • 1
    $\begingroup$ @anon_student : You don't need the full power of the identity theorem. If two rational functions agree on an infinite set, then the difference between them is zero on that set. But the difference is also a rational function. So the numerator of the difference is a polynomial which is identically zero on an infinite set, and so must be the zero polynomial. $\endgroup$ – bryanj Jul 27 '13 at 17:45

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