Help with Proof explanation

Question

I need help in understanding a (topological) proof of Fundamental theorem of algebra.

Here is the Proof:

Suppose $f(z)=a_nz^n+...+a_0$ with $a_0 \neq 0, n \geq1.$ WLOG, assume that $a_n=1.$ We know $f(z)$ is a continuous function $\mathbb{C}\rightarrow \mathbb{C}.$ Further, $$\lim_{z\rightarrow \infty} \frac{z^n}{f(z)}=1,$$ and so for a sufficiently large circle $C_r$ we have $$|z^n - f(z)|\leq \lambda r^n$$ with $0<\lambda < 1, z\in C_r.$ For any $r>0,z^n$ winds $C_r$ around the origin $n$ times.Therefore by the fellow traveller property $f(z)$ will also wind a sufficiently large $C_r$ n times around the origin.

For a small enough r, $f(z)\approxeq a_0$ on $C_r$, so $f(C_r)$ makes a small loop around $a_0$ and will not wind origin at all. Since $f(z)$ is continuous, $f(C_r)$ will depend continuously on $r$. Since $f(C_r)$ has winding number 0 for a small radius r and winds n times around the origin for larger r, there must be intermediate radius $r_1$ with $f(C_{r_1})$ passing through the origin. It then follows that there must be a point $z_0$ on $C_r$ with $f(z_0)=0. $ This proves the theorem.

I do not quite understand the statements that are in bold letters. Thanks.

Answer 1

Here is how I would explain the three "bold statements".

(i) If $z\in C_r$ then $\vert z^n-f(z)\vert =\vert z^n\vert\, \left\vert 1-\frac{f(z)}{z^n}\right\vert= \left\vert 1-\frac{f(z)}{z^n}\right\vert\, r^n$. Since $f(z)/z^n\to 1$ as $\vert z\vert\to \infty$, you have $\left\vert 1-\frac{f(z)}{z^n}\right\vert\leq\frac 12$ if $z\in C_r$ and $r$ is large enough, so that $\vert z^n-f(z)\vert\leq \frac 12\, r^n$.

(ii) Since $a_0\neq 0$, $f(C_r)$ will stay in a disk centred at $a_0$ and not containing $0$ if $r$ is small enough. So $f(C_r)$ will not loop around $0$.

(iii) If you assume that $f(C_r)$ never passes through $0$, then the winding number of $f(C_r)$ will be always defined and will depend continuously on $r$. Since this winding number is equal to $0$ for small $r$ and to $n$ for large $r$, you get a contradiction.

Answer 2

definition: [Index of a function] Let $\gamma$ be a closed curve and $\Omega := \mathbb{C} \setminus \text{supp}(\gamma)$. We define the index of $z \in \Omega$ with respect to $\gamma$ the number $$ \text{Ind}_{\gamma}(z):= \frac{1}{2\pi i} \int_{\gamma} \frac{dw}{w-z}$$

Obviously the index express the winding number of a function around a point which doesn't lie inside its graph.

Fact 1 The index of a function has always integers value.

Fact 2 The index of a continuous function is continuous where defined.

Fact 3 The index of a function is a constant function over the connected components of its domain.

Just because we are dealing with polynomial function, we have that for large $r$ the function behave like $z^{\text{deg}(f)}$. ( $|z|^n > |a_{n−1}z^{n−1} + \cdots + a_0|$). For smaller $r$, it is immediate to see that $f \approx a_0$.

Its is easy to see that, defined $\gamma = r^ne^{in\theta}$ (z^n around the circle with radius r) $$\text{Ind}_{\gamma}(0) = n $$ for large $r$ and so by our assumption, the same is for $f$. If $r$ is sufficient little, the index is $0$ (If $f \approx a_0$ (it is true for small $r$) the function is very similar to a constant, so it doesn't move "much" from the point $a_0$, in particular no windings around the origin: so index 0 for small enough $r$.)

But this is in contrast with Fact 3

In fact the function is: $$r \mapsto \frac{1}{2i\pi}\int_{0}^{2\pi} \frac{f'(re^{i \theta})}{f(re^{i\theta})} d\theta$$ $$\times \times \times \times $$ The main point in this proof is starting by supposing that the polynomial function $f(z)$ has not roots. With this assumption, if we now allow $z$ to trace any closed curve in the complex plane, $f(z)$ will describe a closed curve (same starting and ending point) which never passes through the origin, and so the index of the polynomial function $f$ with respect to the origin can be defined with the above properties. The absurd is that the Index is defined for all $r$ and is continuous but it is not constant and with integer value. An Absurd, so there exists $r_1$ for which the index is not defined, in other words exist a circle $C_{r_1}$ of radius $r_1$ such that $f(C_{r_1})$ "touch" the origin.

For further explanations with pictures look here and here (index, order and winding number sometimes are used like synonyms)