$n$-to-$1$ near zero of holomorphic function

Question

Can someone explain to me why a holomorphic function that grows like a polynomial of degree $n$ is $n$-to-$1$ near it's roots? I keep reading this fact on this site, but I can't find an explanation.

Do you mean, a holomorphic function that has a zero of order $n$ is $n$-to-one near that zero? That follows from the argument principle. — Robert Israel, Jul 17 '14 at 07:00
I think that's what I mean - it's on the answer here http://math.stackexchange.com/questions/29758/entire-1-1-function.
I don't know the argument principle. — , Jul 17 '14 at 07:02
in your cited answer he mentions $n+1$ to one for derivative of order $n$. — Daniel Valenzuela, Jul 17 '14 at 07:06

score 2 · Accepted Answer · answered Jul 17 '14 at 08:05

As indicated in the comments, this is a consequence of the argument principle.

Here's the basic idea. Suppose $f$ has a zero of order $n$ at $a$. Then we may write $$f(z)=(z-a)^n g(z)$$ where $g$ is analytic and non-zero on a neighborhood of $a$. So on a suitably small disc we write $z=re^{i\theta}+a$, whence $$f( re^{i\theta}+a) =r^n e^{in\theta}g( re^{i\theta}+a).$$ The exponential tells us that every full rotation in the domain gives $n$ rotations in the image. Boundedness of $g$ away from zero assures it makes no net contributions: the arguments of its values stay inside an interval of size strictly less than $2\pi$.

score 0 · Answer 2 · answered Jul 17 '14 at 15:23

The proof from the argument principle works like this.

Suppose $\Gamma$ is a positively oriented simple closed contour, let's say a circle, and $f$ is analytic in a domain that includes $\Gamma$ and its interior. Then the sum of the orders of the zeros of $f$ inside $\Gamma$ is $$ \dfrac{1}{2\pi i} \int_\Gamma \dfrac{f'(z)}{f(z)}\; dz$$ If you replace $f(z)$ by $f(z) - w$ you get the sum of the orders of the zeros of $f(z) - w$. Now the crucial point is that in some neighbourhood of $0$ (in fact in the connected component of ${\mathbb C} \backslash f(\Gamma)$ containing $0$) $\frac{1}{2\pi i} \int_{\Gamma} \frac{f'(z)}{f(z)-w}\; dz$ is constant. So for $w$ near $0$, $f(z) = w$ has the same number of solutions (counted by order) inside $\Gamma$ as does $f(z) = 0$. If we avoid points $z$ where $f'(z) = 0$ (of which there are only finitely many inside $\Gamma$) the solutions of $f(z) = w$ all have order $1$.

In particular, if $\Gamma$ contains a single zero $p$ of $f$ and it has order $n$, this says that there is a neighbourhood $V$ of $0$ such that for every $w$ in $V$ except $0$, $f(z) = w$ has exactly $n$ solutions inside $\Gamma$. If $W$ is the part of $f^{-1}(V)$ inside $\Gamma$, then $f$ is $n$-to-one from $W \backslash \{p\}$ to $V \backslash \{0\}$.

Because it's a continuous function (off $f(\Gamma)$) of $w$ that takes on only integer values. — Ted Shifrin, Jul 17 '14 at 18:57

$n$-to-$1$ near zero of holomorphic function

2 Answers2