# Number of zeros equals number of poles

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.

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.

• Couldn't there be poles right on the unit circle? What would happen then? Jan 9, 2014 at 13:45
• 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. Jan 9, 2014 at 13:51

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.

• But how do you see that that integral is $0$? Jan 9, 2014 at 12:08
• @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? Jan 9, 2014 at 13:16
• 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. Jan 9, 2014 at 13:21
• 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? Jan 9, 2014 at 13:34
• It is relevant, since it counts "zeros - poles". Ah, well, beaten by Sourav to the answer. Jan 9, 2014 at 13:37