Fundamental theorem of algebra: an explicit proof closure

The theorem and its prerequisites

Every nonconstant polynomial $p\in\mathbb C[z]$ has a complex root; consequently every degree-$n$ complex polynomial is a product of $n$ linear factors, counted with multiplicity. We prove this conclusion through Goursat, the Cauchy estimate and Liouville, with the prerequisites given below.

Holomorphic means complex differentiable at every point of an open set; entire means holomorphic on all of $\mathbb C$. The elementary foundations used below are real one-variable integration and its chain rule, continuity and differentiation of polynomials, the complex modulus laws, and compactness in the Euclidean plane (including shrinking compact triangles, uniform continuity and a Lebesgue number for a compact cover). These real-analysis and topology foundations are explicitly prerequisites. No complex polynomial-root theorem, maximum-modulus theorem, analytic power-series representation, identity theorem, residue theorem or argument principle is assumed.

Complex differentiability implies continuity, since $u(z+h)-u(z)=h[(u(z+h)-u(z))/h]\to0$. Expanding a product difference quotient gives $(uv)'=u'v+uv'$. Where $v(z)\ne0$,

$$\frac{1/v(z+h)-1/v(z)}h=-\frac{v(z+h)-v(z)}{h\,v(z+h)v(z)}\longrightarrow-\frac{v'(z)}{v(z)^2}.$$

Thus products and nonvanishing quotients of holomorphic functions are holomorphic. In particular $f(z)/(z-a)$ is holomorphic away from $a$, and a root-free polynomial has an entire reciprocal. These facts use difference quotients, not a polynomial-root or regularity theorem.

1. An integral records both a path and its orientation

A piecewise $C^1$ curve is a continuous map $\gamma:[a,b]\to\mathbb C$ that is continuously differentiable on finitely many subintervals, with one-sided derivatives at the joins. It is closed when $\gamma(a)=\gamma(b)$. Its length is $\ell(\gamma)=\int_a^b|\gamma'(t)|\,dt$. A contour integral of continuous $f$ along it is

$$\int_\gamma f(z)\,dz=\int_a^b f(\gamma(t))\gamma'(t)\,dt,$$

with real and imaginary parts integrated separately. The differential $dz$ includes the direction of motion; replacing it by $|dz|$ would define a different integral.

A piecewise $C^1$ increasing change of parameter preserving endpoints leaves the integral unchanged by substitution. Reversal changes its sign. Concatenating two compatible curves adds their integrals. These statements also allow zero-speed pieces. They do not require the curve to be injective.

Estimation lemma. If $|f|\le M$ on the curve, then

$$\left|\int_\gamma f(z)\,dz\right|\le M\ell(\gamma).$$

Indeed, the modulus of a real-parameter integral is at most the integral of the modulus, and $|f(\gamma)\gamma'|\le M|\gamma'|$. This simple inequality is the main tool for showing that auxiliary arcs contribute nothing in a limit.

Worked example. On the positively oriented circle $z=a+re^{it}$, $0\le t\le2\pi$, with $r>0$,

$$\int_\gamma(z-a)^n\,dz=\begin{cases}2\pi i,&n=-1,\\0,&n\in\mathbb Z,\ n\ne-1.\end{cases}$$

For $n=-1$ the integrand times $dz$ is $i\,dt$. Otherwise it is $ir^{n+1}e^{i(n+1)t}\,dt$, whose integral vanishes. The answer for $n=-1$ is independent of radius, but changes sign if orientation is reversed.

2. Primitives turn path integrals into endpoint differences

A primitive of $f$ on an open set $U$ is a holomorphic function $F$ there with $F'=f$. The real chain rule applied on each smooth piece gives

$$\int_\gamma f(z)\,dz=F(\gamma(b))-F(\gamma(a))$$

for every contour in $U$. Hence a function with a primitive has zero integral around every closed contour in $U$.

Conversely, on a domain $U$, suppose continuous $f$ has zero integral around every closed contour. Open connected subsets of the plane are polygonally connected. Fix $a\in U$ and define $F(z)$ as the integral from $a$ to $z$ along any polygonal path in $U$. Closed-path cancellation makes it path independent. For sufficiently small $h$, the final straight segment from $z$ to $z+h$ stays in $U$, and

$$\frac{F(z+h)-F(z)}h=\int_0^1 f(z+th)\,dt\longrightarrow f(z)$$

by continuity. Thus $F'=f$. The same construction works if polygonal closed paths alone have zero integrals; the resulting primitive then handles all contours.

3. Goursat's triangle argument: no continuity of the derivative assumed

Theorem. If $f$ is holomorphic on an open set containing a closed triangle $T$, then its integral around the positively oriented boundary of $T$ is zero.

Proof. Subdivide $T$ by joining side midpoints into four triangles. Interior edge integrals cancel, so the outer integral $I$ is the sum of four child integrals. At least one child has integral modulus at least $|I|/4$. Repeat inside such a child, obtaining nested closed triangles $T_n$, diameter $D/2^n$, perimeter $P/2^n$, and

$$|I|\le4^n\left|\int_{\partial T_n}f\,dz\right|.$$

Compactness and shrinking diameters give one common point $z_0$. Differentiability at $z_0$ says

$$f(z)=f(z_0)+f'(z_0)(z-z_0)+(z-z_0)\varepsilon(z),\qquad\varepsilon(z)\to0.$$

The constant and linear terms have primitives, so contribute zero on the triangle boundary. For any $\eta>0$, eventually $|\varepsilon|\le\eta$ throughout $T_n$. Since $z_0\in T_n$, the estimation lemma gives

$$\left|\int_{\partial T_n}f\,dz\right|\le\eta(D/2^n)(P/2^n).$$

Thus $|I|\le\eta DP$. As $\eta$ is arbitrary, $I=0$. No continuity of $f'$ was used. ∎

Corollary: a primitive on each disk. If $f$ is holomorphic on a disk $D$, fix its centre $a$ and set $F(z)=\int_{[a,z]}f$. For $z,z+h\in D$ sufficiently close, the triangle with vertices $a,z,z+h$ lies in $D$. The triangle theorem gives $F(z+h)-F(z)=\int_{[z,z+h]}f$. The limit calculation of §2 yields $F'=f$. Consequently all closed-contour integrals in the disk vanish. The same argument works on any open convex set.

4. Homotopy and what “no hole” actually means

A fixed-endpoint homotopy between paths $\gamma_0,\gamma_1$ is a continuous map $H:[0,1]^2\to U$ with $H(s,0)=\gamma_0(s)$, $H(s,1)=\gamma_1(s)$, and $H(0,t),H(1,t)$ constant in $t$. It deforms one path into the other without leaving $U$. A domain is simply connected if every closed path can be contracted to its basepoint by such a homotopy.

Homotopy invariance. If $f$ is holomorphic on $U$, and two piecewise $C^1$ paths in $U$ are fixed-endpoint homotopic there, their integrals agree.

Proof using only local primitives. The compact image of $H$ is covered by finitely many disks on which $f$ has primitives. By the Lebesgue-number property and uniform continuity, subdivide the parameter square into a fine rectangular grid so that the image of each closed cell lies in one such disk. Join the images of adjacent grid vertices by straight segments. For each cell these segments lie in its disk because disks are convex. Its polygonal boundary integral is zero by its primitive. Sum over cells: internal segments cancel with their reversals. On the bottom and top edges, each original path subarc and its replacing straight segment lie in the same disk and have the same endpoints, hence the same integral by the local primitive. The side segments vanish because endpoints were fixed. What remains is the bottom integral minus the top integral. This proves the claim even if the homotopy's intermediate paths are not differentiable. ∎

The open set for the circular deformation

In the Cauchy proof, let $U$ be the original open holomorphy neighbourhood of $\overline{D(c,R)}$, and let $d=|a-c|<R$. Choose $0<\varepsilon<R-d$. The correct holomorphy set for $g(z)=f(z)/(z-a)$ is $U\setminus\{a\}$, including the outer circle; the open disk $D(c,R)\setminus\{a\}$ alone would exclude that circle. This open set includes the boundary circle, and the following inequalities justify the deformation inside it.

Put $c_t=(1-t)c+ta$, $r_t=(1-t)R+t\varepsilon$, and $H(s,t)=c_t+r_te^{2\pi i s}$. For $0\le s,t\le1$,

$$|H(s,t)-c|\le R-t(R-d-\varepsilon)\le R,$$ $$|H(s,t)-a|\ge r_t-|c_t-a|=(1-t)(R-d)+t\varepsilon>0.$$

Thus the whole compact homotopy image lies in $U\setminus\{a\}$. The grid/local-primitive proof works for this free homotopy of closed loops: $H(0,t)=H(1,t)$, so the two polygonal side tracks are identical, have opposite boundary orientations and cancel. They need not be constant. The top and bottom circles are piecewise $C^1$; no differentiability of the intervening homotopy is needed. This supplies exactly the domain and moving-basepoint step used next.

1. A boundary integral determines an interior value

For $c\in\mathbb C$ and $R>0$, orient the circle $C:|z-c|=R$ counterclockwise. Suppose $f$ is holomorphic on an open neighbourhood of the closed disk $\overline{D(c,R)}$. For every $a$ with $|a-c|<R$,

$$f(a)=\frac1{2\pi i}\int_C\frac{f(z)}{z-a}\,dz.$$

The denominator is never zero on $C$. The formula is an interior-value theorem; it does not assert this value for an exterior point or a point on the contour.

Proof. Choose $\varepsilon>0$ with $|a-c|+\varepsilon<R$. The outer circle deforms to the small circle $C_\varepsilon:|z-a|=\varepsilon$ without passing through $a$: use centres $(1-t)c+ta$ and radii $(1-t)R+t\varepsilon$. Each moving circle encloses $a$ and stays in the original disk. Apply the grid/local-primitive proof of C02 to $f(z)/(z-a)$ on the punctured disk. Although the moving basepoint is not fixed, its two side tracks are identical and cancel, so closed-loop integrals are unchanged. Hence

$$\int_C\frac{f(z)}{z-a}\,dz=\int_{C_\varepsilon}\frac{f(z)}{z-a}\,dz =2\pi if(a)+\int_{C_\varepsilon}\frac{f(z)-f(a)}{z-a}\,dz.$$

The final term has modulus at most $2\pi\sup_{|z-a|=\varepsilon}|f(z)-f(a)|$, which tends to zero by continuity at $a$. This proves the formula. ∎

This proof does not presume that $(f(z)-f(a))/(z-a)$ has a holomorphic extension at $a$; that fact is a consequence of the regularity obtained below.

The first derivative, with a uniform bound

Only the first Cauchy derivative estimate is needed for Liouville. For an interior point $a$ of a fixed circle $C$, set $\delta=\min_{z\in C}|z-a|>0$. When $0<|h|<\delta/2$, subtraction of the two Cauchy formulas gives the difference quotient kernel

$$\frac1h\left(\frac1{z-a-h}-\frac1{z-a}\right)=\frac1{(z-a-h)(z-a)}.$$

Its difference from $1/(z-a)^2$ is

$$\frac{h}{(z-a-h)(z-a)^2},\qquad \sup_{z\in C}\left|\frac{h}{(z-a-h)(z-a)^2}\right|\le\frac{2|h|}{\delta^3}.$$

If $M=\max_C|f|$ and $L=\ell(C)$, the estimation lemma bounds the error after division by $2\pi i$ and integration by $ML|h|/(\pi\delta^3)$, which tends to zero. Hence $f'(a)=(2\pi i)^{-1}\int_C f(z)/(z-a)^2\,dz$. For a circle of radius $r$ centred at $a$, this gives $|f'(a)|\le M_r/r$. Repeating the same argument gives the higher-order formula below; Liouville uses only the first-order estimate.

2. All higher derivatives and their bounds

For $a$ in the open disk the distance from $a$ to the contour is positive. On a sufficiently small disk around $a$, difference quotients of the kernel $1/(z-a)$ converge uniformly in $z\in C$ to $1/(z-a)^2$. The estimation lemma permits passing that limit under the contour integral. Repeating this argument yields

$$f^{(n)}(a)=\frac{n!}{2\pi i}\int_C\frac{f(z)}{(z-a)^{n+1}}\,dz,\qquad n=0,1,2,\ldots.$$

The uniform kernel argument proves the existence of these derivatives; it does not assume $f$ was initially $C^\infty$. The same formula shows they are continuous locally because the denominators stay uniformly separated from zero.

On a circle centred at $a$ of radius $r$, put $M_r=\max_{|z-a|=r}|f(z)|$. Length estimation gives the Cauchy estimates

$$|f^{(n)}(a)|\le\frac{n!M_r}{r^n}.$$

The radius must belong to a disk on which the function is holomorphic up to a neighbourhood of its boundary. Taking a larger radius through a singularity is invalid.

Liouville and the polynomial-root theorem

For an entire function, the preceding estimate is available at every radius. Applying it to a bounded entire function gives Liouville; applying Liouville to a polynomial reciprocal proves root existence.

Liouville's theorem. A bounded entire function is constant. If $|f|\le M$ on $\mathbb C$, apply the $n=1$ estimate on a disk of arbitrary radius $r$ about any $a$: $|f'(a)|\le M/r$. Sending $r\to\infty$ gives $f'=0$, so the function is constant on the connected plane.

Fundamental theorem of algebra. Every nonconstant complex polynomial has a complex root. If $p$ had no root, $1/p$ would be entire. Its leading term implies $|p(z)|\to\infty$ as $|z|\to\infty$, so $1/p$ is bounded outside a large disk; continuity gives boundedness on the remaining compact disk. Liouville makes $1/p$ constant, contradicting the degree of $p$. Repeated division by linear factors then factors every degree-$n$ polynomial into exactly $n$ linear factors counted with multiplicity.

Elementary details of the polynomial contradiction

For completeness, a function $F$ with $F'=0$ on $\mathbb C$ is constant: along $\gamma(t)=a+t(b-a)$ the real and imaginary parts of $F(\gamma(t))$ have derivative zero, so the real mean-value theorem gives $F(a)=F(b)$.

Write $p(z)=a_nz^n+\sum_{k<n}a_kz^k$ with $n\ge1$, $a_n\ne0$, and $S=\sum_{k<n}|a_k|$. For $|z|\ge R=\max(1,2S/|a_n|)$,

$$|p(z)|\ge |a_n||z|^n-S|z|^{n-1}\ge\frac{|a_n|}{2}|z|^n.$$

If $p$ had no root, its reciprocal would be continuous everywhere and holomorphic, since

$$\frac{1/p(z+h)-1/p(z)}h=-\frac{p(z+h)-p(z)}{h\,p(z+h)p(z)}\longrightarrow-\frac{p'(z)}{p(z)^2}.$$

The displayed growth bound makes $1/p$ bounded outside the radius-$R$ disk. It is bounded on the remaining closed disk by compactness and continuity. Liouville makes $1/p$ a constant, which is nonzero, forcing $p$ to be constant, a contradiction. Thus a root $\alpha$ exists.

Finally $z^k-\alpha^k=(z-\alpha)\sum_{j=0}^{k-1}z^{k-1-j}\alpha^j$ gives $p(z)-p(\alpha)=(z-\alpha)q(z)$, where $q$ has degree $n-1$ and leading coefficient $a_n$. Apply the root theorem to $q$ if it is nonconstant and induct on degree. After exactly $n$ divisions the remaining factor is the nonzero constant $a_n$. Degree and the product identity give exactly $n$ linear factors with multiplicity. This also proves the complex splitting assertion needed by finite-dimensional spectral or Jordan arguments; it does not assert splitting over an arbitrary real or other ground field.