# An unpacking of Hurwitz’s theorem in complex analysis

Originally published at 狗和留美者不得入内. You can comment here or there.

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 for every small enough $\rho > 0$, there is $k$ large such that $f_k(z)$ 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 in $|z - z_0| < \rho$, $\rho$ 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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