An unpacking of Hurwitz’s theorem in complex analysis

Let’s first state it.

Theorem (Hurwitz’s theorem). Suppose \{f_k(z)\} is a sequence of analytic functions on a domain D that converges normally on D to f(z), and suppose that f(z) has a zero of order N at z_0. Then there exists \rho > 0 such that for k large, f_k(z) has exactly N zeros in the disk \{|z - z_0| < \rho\}, counting multiplicity, and these zeros converge to z_0 as k \to \infty.

As a refresher, normal convergence on D is convergence uniformly on every closed disk contained by it. We know that the argument principle comes in handy for counting zeros within a domain. That means

The number of zeros inside |z - z_0| = \rho, \rho arbitrarily small goes to the number of zeros inside the same circle of f, provided that

\frac{1}{2\pi i}\int_{|z - z_0| = \rho} \frac{f'_k(z)}{f_k(z)}dz \longrightarrow \frac{1}{2\pi i}\int_{|z - z_0| = \rho} \frac{f'(z)}{f(z)}dz.

To show that boils down to a few technicalities. First of all, let \rho > 0 be sufficiently small that the closed disk \{|z - z_0| \leq \rho\} is contained in D, with f(z) \neq 0 inside it everywhere except for z_0. Since f_k(z) converges to f(z) uniformly inside that closed disk, f_k(z) is not zero on its boundary, the domain integrated over, for sufficiently large k. Further, since f_k \to f uniformly, so does f'_k / f_k \to f' / f, so we have condition such that convergence is preserved on application of integral to the elements of the sequence and to its convergent value. With \rho arbitrarily small, the zeros of f_k(z) must accumulate at z_0.

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