Fundamental Theorem of Algebra I

Let $\displaystyle P(z)=a_nz^n+a_{n-1}z^{n-1}+\cdots+a_0,\quad a_n\neq0,$ where $n\ge1$ . We shall prove that $P$ has exactly $n$ complex roots, counted with multiplicity.

The central observation is that the degree- $n$ term governs the polynomial at large scale. When $|z|$ is large, every lower-degree monomial is smaller than $z^n$ by at least one factor of $1/|z|$ . Thus, on a sufficiently large circle, the curve traced by $P(z)$ is a small perturbation of the curve traced by $a_nz^n$ . The latter goes around the origin exactly $n$ times as $z$ makes one positive circuit around the circle. We will show that the lower-order terms cannot alter this integer winding number.

Counting Zeroes

Let $\displaystyle C_R:={z\in\mathbb C:|z|=R}$ , oriented counterclockwise, and suppose for the moment that $P$ has no zero on $C_R$ . Consider

$\displaystyle N(R):=\frac{1}{2\pi i}\int_{C_R}\frac{P'(z)}{P(z)} dz.$

This integral counts the zeros of $P$ inside the circle, with their multiplicities. Indeed, let $\alpha$ be a root of multiplicity $m.$ In a neighborhood of $\alpha,$ one can factor

$\displaystyle P(z)=(z-\alpha)^m h(z),\quad h(\alpha)\neq0.$

Taking a logarithmic derivative gives

$\displaystyle \frac{P'(z)}{P(z)}=\frac{m}{z-\alpha}+\frac{h'(z)}{h(z)}.$

The second term is holomorphic near $\alpha,$ while the first has residue $m.$ Thus each root contributes precisely its multiplicity to the residue theorem, and therefore

$\displaystyle N(R)=\sum_{{\alpha:P(\alpha)=0, ~ |\alpha|<R}}m_\alpha.$

In particular, $N(R)$ is a nonnegative integer. The fundamental theorem of algebra will follow once we show that $N(R)=n$ for all sufficiently large $R.$

There is also a useful geometric interpretation. Along the contour $C_R,$ the form $P'(z)/P(z) dz$ is the infinitesimal change of a logarithm of $P(z).$ Hence $N(R)$ is the winding number of the closed curve $P(C_R)$ about the origin. The residue calculation says that this winding number is exactly the total multiplicity of the roots enclosed by $C_R.$

Estimation

Factor out the leading term and write $\displaystyle P(z)=a_nz^n\Big(1+E(z)\Big),$ where

$\displaystyle E(z):=\frac{a_{n-1}}{a_n}\frac1z+\frac{a_{n-2}}{a_n}\frac1{z^2}+\cdots+\frac{a_0}{a_n}\frac1{z^n}.$

The function $E(z)$ is the relative error made by replacing $P(z)$ with its leading term $a_nz^n.$ This is the right error to study: although the difference $\displaystyle P(z)-a_nz^n$ may itself be large when $|z|$ is large, it is small compared with the dominant quantity $a_nz^n.$

Write $E(z)=\sum_{j=1}^n b_jz^{-j}, ~b_j:=\frac{a_{n-j}}{a_n},$ and define $A:=\sum_{j=1}^n|b_j|,~ B:=\sum_{j=1}^n j|b_j|.$ When $|z|=R\ge1,$ every negative power $z^{-j}$ has size at most $R^{-1}.$ Hence

$\displaystyle |E(z)|\le\sum_{j=1}^n|b_j|R^{-j}\le\frac{A}{R}.$

Likewise, $E'(z)=-\sum_{j=1}^n jb_jz^{-j-1},$ so

$\displaystyle |E'(z)|\le\sum_{j=1}^n j|b_j|R^{-j-1}\le\frac{B}{R^2}.$

Choose $R$ larger than enough compared $1,$ $2A,$ and $2B.$ Then, throughout the circle $C_R,$ $|E(z)|<\frac12.$ Consequently,

$\displaystyle |1+E(z)|\ge1-|E(z)|>\frac12.$

In particular, $\displaystyle 1+E(z)\neq0$ on $C_R,$ and therefore

$\displaystyle P(z)=a_nz^n\Big(1+E(z)\Big)\neq0\quad\text{for }|z|=R.$

Thus the contour integral defining $N(R)$ is valid.

Main term

Differentiate the factorization $\displaystyle P(z)=a_nz^n\Big(1+E(z)\Big).$ Since the logarithmic derivative of a product is the sum of the logarithmic derivatives of its factors,

$\displaystyle \frac{P'(z)}{P(z)}=\frac{n}{z}+\frac{E'(z)}{1+E(z)}.$

The first term is the contribution of the leading monomial $a_nz^n.$ Its contour integral is exact:

$\displaystyle \frac{1}{2\pi i}\int_{C_R}\frac{n}{z} dz=n.$

Therefore

$\displaystyle N(R)=n+\frac{1}{2\pi i}\int_{C_R}\frac{E'(z)}{1+E(z)} dz.$

Thus all that remains is to show that the correction term cannot change the integer $n.$

Error term

On $C_R,$ the preceding bounds give

$\displaystyle \Big|\frac{E'(z)}{1+E(z)}\Big|\le\frac{|E'(z)|}{|1+E(z)|}\le\frac{B/R^2}{1/2}=\frac{2B}{R^2}.$

The contour $C_R$ has length $2\pi R.$ Hence

$\displaystyle \Big|N(R)-n\Big|\le\frac{1}{2\pi}(2\pi R)\frac{2B}{R^2}=\frac{2B}{R}\to 0.$

But $N(R)$ is an integer, so $N(R)-n =O(\frac{1}{R})$ is an integer as well and it has to be $0.$ Thus

$\displaystyle N(R)=n.$

We have proved that every sufficiently large disk contains exactly $n$ roots of $P,$ counted with multiplicity. Therefore every nonconstant polynomial of degree $n$ has exactly $n$ complex roots, counted with multiplicity.

The final estimate is a quantitative way of expressing a topological stability principle. On the large circle $C_R,$ the factor $\displaystyle 1+E(z)$ lies in the disk $\{w\in\mathbb C:|w-1|<\frac12 \}.$ This disk does not contain the origin. Thus the curve $\displaystyle 1+E(z)$ cannot wind around $0.$ Equivalently, the deformation $a_nz^n(1+tE(z)),\quad 0\le t\le1,$ never vanishes on $C_R,$ because $|1+tE(z)|\ge1-t|E(z)|>\frac12.$ As $t$ moves from $0$ to $1,$ this deformation continuously changes the leading monomial $a_nz^n$ into the full polynomial $P(z),$ without ever allowing the boundary curve to pass through the origin. Therefore their winding numbers are the same. The polynomial $a_nz^n$ winds exactly $n$ times around $0,$ so $P$ must do the same. The contour estimate above proves this stability numerically: there is neglible correction to the leading winding number. At large distance, the lower-degree terms may bend the curve $P(C_R),$ but they are too small relative to $a_nz^n$ to pull that curve across the origin or change its total winding. The degree of the polynomial is therefore remembered globally as the total number of its complex zeros.