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The following is an old qualifying exam problem I cannot solve:

Let $f$ be a meromorphic function (quotient of two holomorphic functions) on an open neighborhood of the closed unit disk. Suppose that the imaginary part of $f$ does not have any zeros on the unit circle, then the number of zeros of $f$ in the unit disk equals the number of poles of $f$ in the unit disk.

It seems this problem begs Rouche's theorem, but I cannot seem to apply it correctly.

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2 Answers 2

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The imaginary part of $f$ is never zero on the unit circle and hence $f(\partial \mathbb{D})$ never becomes zero that is never crosses the Real axis. This in turn means that as $z$ traverses counterclockwise along the unit circle, $f(z)$ never makes a complete rotation around the origin.

This is another form of the argument principle which states that: the number of zeros - the number of poles of a meromorphic $f$ inside a simple closed curve ($f$ cannot be zero on the curve) = $\frac{1}{2\pi } \times$(change in $\arg{f(z)}$ as z traverses around the curve counterclockwise) = number of times $f$ winds around zero counterclockwise.

Here as we saw, $f$ never winds around zero while traversing the unit circle, hence the number of zeros equals the number of poles of $f$ inside the unit disk.

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  • $\begingroup$ Couldn't there be poles right on the unit circle? What would happen then? $\endgroup$
    – user1337
    Jan 9, 2014 at 13:45
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    $\begingroup$ Firstly there are a finite number of zeros of $f$ inside the unit disk. If we divide by the factors and define $g = \dfrac{f}{\prod_{n=1}^k(z-z_n)}$, then $g$ is zero-free in the unit disk and meromorphic. Then $1/g$ is holomorphic in a neighborhood of the closed unit disk with zeros at the poles of $f$. Zeros cannot accumulate anywhere in the interior of the domain and the unit circle is in the interior. Thus $f$ only has a finite set of poles in the unit disk along with a finite set of zeros. $\endgroup$
    – Sourav D
    Jan 9, 2014 at 13:51
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This is not quite Rouche though everything in the end boils down to the Cauchy integral formula. To count zeros and poles you pass to the logarithmic derivative i.e. $\frac{f'}{f}$ and then integrate over the unit circle (which you are allowed to do since $f$ does not vanish there by hypothesis). Why you can take a continuous branch of the log is explained in the answer by @Sourav.

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  • $\begingroup$ But how do you see that that integral is $0$? $\endgroup$
    – Daniel Fischer
    Jan 9, 2014 at 12:08
  • $\begingroup$ @DanielFischer, I see no reason why it should be. The holomorphic function $f(z)=z$ has one zero in the unit disk and no poles there. I was just pointing out that the right way of looking at this is from the viewpoint of the logarithmic derivative. What's your guess for the missing hypothesis? $\endgroup$
    – Mikhail Katz
    Jan 9, 2014 at 13:16
  • $\begingroup$ There is no missing hypothesis, it's a variant of (the proof of) Rouché's theorem. Since you didn't deliberately leave that to the OP to figure out, I guess I'll post an answer explaining it. $\endgroup$
    – Daniel Fischer
    Jan 9, 2014 at 13:21
  • $\begingroup$ Thanks, sorry, I see the problem is subtler than I thought. I should probably delete my answer. Is the log-derivative relevant at all here? $\endgroup$
    – Mikhail Katz
    Jan 9, 2014 at 13:34
  • $\begingroup$ It is relevant, since it counts "zeros - poles". Ah, well, beaten by Sourav to the answer. $\endgroup$
    – Daniel Fischer
    Jan 9, 2014 at 13:37

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