Let \(U\) be an open set in \(\mathbb{C}\) containing the closed unit disk \(\overline{D}\). Suppose \(f\) is a meromorphic function on \(U\) such that \(f(\partial D) \subset \mathbb{R}^{+}\). (In particular, \(f\) has no zeros or poles on \(\partial D\).) Show that \(f\) has the same number of zeros as poles in \(D\), counting multiplicities.
Solution
1. The logarithmic derivative has integral zero around the unit circle.
Since \(f\) has no pole on \(\partial D\), it is holomorphic on a neighborhood of that circle. Put \(c(t)=f(e^{it})\) for \(0\leq t\leq2\pi\). This is a continuously differentiable function with values in the positive real numbers, and \(c(0)=c(2\pi)\). The ordinary real logarithm is therefore defined along its entire image and has derivative \(c'/c\). By the chain rule, \[ \int_{|z|=1}\frac{f'(z)}{f(z)}\,dz =\int_0^{2\pi}\frac{c'(t)}{c(t)}\,dt =\log c(2\pi)-\log c(0)=0. \] The circle is oriented counterclockwise. This uses only a logarithm along the boundary image, not a holomorphic logarithm at any interior zero or pole.
2. The argument principle equates the zero and pole counts.
The function is meromorphic on a neighborhood of the closed disk and has neither zeros nor poles on its boundary. It is not identically zero on the component containing the disk, since its boundary values are positive. Its zeros and poles in the closed disk consequently form a finite set, with finite orders, by their local meromorphic factorizations and compactness [@SS03]. The argument principle gives \[ N-P=\frac1{2\pi i}\int_{|z|=1}\frac{f'}f\,dz=0, \] where \(N\) is the sum of the zero orders and \(P\) the sum of the pole orders in \(D\) [@SS03]. Thus \(N=P\), exactly the requested equality with multiplicities.