Polynomials, Rouche's theorem and index of vector fields In the proof of Rouche's theorem I saw in a book, there are two points I failed to understand, or failed to prove myself. (if you aren't familiar with the theorem, please try to look at the two statements here below and explain them regardless of the theorem):


*

*"Consider the complex plane. It's not difficult to show that the index of a curve $C$ with respect to a vector field $v$ is equal to the sum of indices of singular points, i.e., those at which $v(z) = 0$."

*"If $f(z)$ is a polynomial, and let $v(z) = f(z)$, where $v$ is a vector field, then the index of the singular point $z_0$ is equal to the multiplicity of the root $z_0$ of $f$."
I am having trouble understanding this; can anyone please help?
 A: *

*If by index of the curve C they mean the integral


$$ \frac{1}{2 \pi i} \int_\gamma \frac{v'(z)}{v(z)} dz ,$$
which is what I am guessing that they mean, then this follows from the residue theorem applied to the complex valued function $v(z)$.


*The index of a vector field at $z_0$ is defined to be the winding number you get for taking a small circle about $z_0$ of winding number $1$ and computing the winding number of its image about $0$. This can be computed as:


$$ \frac{1}{2 \pi i} \int_\gamma \frac{v'(z)}{v(z)} dz ,$$
where $\gamma$ is a parametrization of a sufficiently small circle which winds around $z_0$ once. For $v(z) = f(z)$ and $f(z_0) = 0$, we can write $f(z) = (z-z_0)^n g(z)$, where $n$ is maximal such that $g(z_0) \neq 0$. Then,
$$ f'(z) = n (z-z_0)^{n-1} g(z) + (z-z_0)^n g'(z) $$
and
$$ \frac{f'(z)}{f(z)} = \frac{n}{z-z_0} + (z-z_0)^n \frac{g'(z)}{g(z)}.$$
The second term above is analytic in this region if we make $\gamma$ a small enough circle so that $g(z)$ is non-zero in the interior. Thus, when we integrate we just get that the index is
$$ \frac{1}{2 \pi i} \int_\gamma \frac{n}{z - z_0} dz = n ,$$
which is the multiplicity of $z_0$.
