# Hurwitz's theorem (complex analysis)

In mathematics and in particular the field of complex analysis, Hurwitz's theorem is a theorem associating the zeroes of a sequence of holomorphic, compact locally uniformly convergent functions with that of their corresponding limit. The theorem is named after Adolf Hurwitz.

## Statement

Let {f$_k$} be a sequence of holomorphic functions on a connected open set G that converge uniformly on compact subsets of G to a holomorphic function f which is not constantly zero on G. If f has a zero of order m at z$_0$ then for every small enough ρ > 0 and for sufficiently large k ∈ N (depending on ρ), f$_k$ has precisely m zeroes in the disk defined by |z − z$_0$| < ρ, including multiplicity. Furthermore, these zeroes converge to z$_0$ as k → ∞.

## Remarks

The theorem does not guarantee that the result will hold for arbitrary disks. Indeed, if one chooses a disk such that f has zeroes on its boundary, the theorem fails. An explicit example is to consider the unit disk D and the sequence defined by
    $f_{n}(z)=z-1+{\frac {1}{n}},\qquad z\in \mathbb {C}$
which converges uniformly to f(z) = z − 1. The function f(z) contains no zeroes in D; however, each f$_n$ has exactly one zero in the disk corresponding to the real value 1 − (1/n).

## Applications

Hurwitz's theorem is used in the proof of the Riemann mapping theorem, and also has the following two corollaries as an immediate consequence:
* Let G be a connected, open set and {f$_n$} a sequence of holomorphic functions which converge uniformly on compact subsets of G to a holomorphic function f. If each f$_n$ is nonzero everywhere in G, then f is either identically zero or also is nowhere zero.
* If {f$_n$} is a sequence of univalent functions on a connected open set G that converge uniformly on compact subsets of G to a holomorphic function f, then either f is univalent or constant.

## Proof

Let f be an analytic function on an open subset of the complex plane with a zero of order m at z$_0$, and suppose that {f$_n$} is a sequence of functions converging uniformly on compact subsets to f. Fix some ρ > 0 such that f(z) ≠ 0 in 0 < |z − z$_0$| ≤ ρ. Choose δ such that |f(z)| > δ for z on the circle |z − z$_0$| = ρ. Since f$_k$(z) converges uniformly on the disc we have chosen, we can find N such that |f$_k$(z)| ≥ δ/2 for every k ≥ N and every z on the circle, ensuring that the quotient f$_k$′(z)/f$_k$(z) is well defined for all z on the circle |z − z$_0$| = ρ. By Weierstrass's theorem we have $f_{k}'\to f'$ uniformly on the disc, and hence we have another uniform convergence:
    ${\frac {f_{k}'(z)}{f_{k}(z)}}\to {\frac {f'(z)}{f(z)}}.$
Denoting the number of zeros of f$_k$(z) in the disk by N$_k$, we may apply the argument principle to find
    $m={\frac {1}{2\pi i}}\int _{\vert z-z_{0}\vert =\rho }{\frac {f'(z)}{f(z)}}\,dz=\lim _{k\to \infty }{\frac {1}{2\pi i}}\int _{\vert z-z_{0}\vert =\rho }{\frac {f'_{k}(z)}{f_{k}(z)}}\,dz=\lim _{k\to \infty }N_{k}$
In the above step, we were able to interchange the integral and the limit because of the uniform convergence of the integrand. We have shown that N$_k$ → m as k → ∞. Since the N$_k$ are integer valued, N$_k$ must equal m for large enough k.