Fundamental Theorem of Algebra
Theorem
Every non-constant polynomial with coefficients in $\C$ has a root in $\C$.
Proof using Algebraic Topology
Let $\map p z$ be a polynomial in $\C$:
- $\map p z = z^m + a_1 z^{m-1} + \cdots + a_m$
where not all of $a_1, \ldots, a_m$ are zero.
Define a homotopy:
- $\map {p_t} z = t \map p z + \left({1-t}\right) z^m$
Then:
- $\dfrac {\map {p_t} z} {z^m} = 1 + t \paren {a_1 \dfrac 1 z + \cdots + a_m \dfrac 1 {z^m} }$
The terms in the parenthesis go to $0$ as $z \to \infty$.
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Therefore, there is an $r \in \R_{>0}$ such that:
- $\forall z \in \C: \size z = r: \forall t \in \closedint 0 1: \map {p_t} z \ne 0$
Hence the homotopy:
- $\dfrac {p_t} {\size {p_t} }: S \to \Bbb S^1$
is well-defined for all $t$.
This shows that for any complex polynomial $\map p z$ of order $m$, there is a circle $S$ of sufficiently large radius in $\C$ such that $\dfrac {\map p z} {\size {\map p z}}$ and $\dfrac {z^m} {\size {z^m} }$ are freely homotopic maps $S \to \Bbb S^1$.
Hence $\dfrac {\map p z} {\size {\map p z} }$ must have the same degree of $\paren {z / r}^m$, which is $m$.
When $m > 0$, that is $p$ is non-constant, this result and the Extendability Theorem for Intersection Numbers imply $\dfrac {\map p z} {\size {\map p z} }$ does not extend to the disk $\Int S$, implying $\map p z = 0$ for some $z \in \Int S$.
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$\blacksquare$
Proof using Liouville's Theorem
Let $\map P z = a_n z^n + \dots + a_1 z + a_0, \ a_n \ne 0$.
Aiming for a contradiction, suppose that $\map P z$ is not zero for any $z \in \C$.
It follows that $1 / \map P z$ must be entire; and is also bounded in the complex plane.
In order to see that it is indeed bounded, we recall that $\exists R \in \R_{>0}$ such that:
$\cmod {\dfrac 1 {\map P z} } < \dfrac 2 {\cmod {a_n} R^n}, \text{whenever} \ \cmod z > R$.
Hence, $1 / \map P z$ is bounded in the region outside the disk $\cmod z \le R$.
However, $1 / \map P z$ is continuous on that closed disk, and thus it is bounded there as well.
Furthermore, we observe that $1 / \map P x$ must be bounded in the whole plane.
Through Liouville's Theorem, $1 / \map P x$, and thus $\map P x$, is constant.
This is a contradiction.
$\blacksquare$
Proof using Cauchy-Goursat Theorem
Let $p: \C \to \C$ be a complex, non-constant polynomial.
Aiming for a contradiction, suppose that $\map p z \ne 0$ for all $z \in \C$.
Now consider the closed contour integral:
- $\ds \oint \limits_{\gamma_R} \frac 1 {z \cdot \map p z} \rd z$
where $\gamma_R$ is a circle with radius $R$ around the origin.
By Derivative of Complex Polynomial, the polynomial $z \cdot \map p z$ is holomorphic.
Since $\map p z$ is assumed to have no zeros, the only zero of $z \cdot \map p z$ is $0 \in \C$.
Therefore by Reciprocal of Holomorphic Function $\dfrac 1 {z \cdot \map p z}$ is holomorphic in $\C \setminus \set 0$.
Hence the Cauchy-Goursat Theorem implies that the value of this integral is independent of $R > 0$.
On the one hand, one can calculate the value of this integral in the limit $R \to 0$ (or use Cauchy's Residue Theorem), using the parameterization $z = R e^{i \phi}$ of $\gamma_R$:
\(\ds \lim \limits_{R \mathop \to 0} \oint \limits_{\gamma_R} \frac 1 {z \cdot \map p z} \rd z\) | \(=\) | \(\ds \lim \limits_{R \mathop \to 0} \int \limits_0^{2 \pi} \frac 1 {R e^{i \phi} {\map p {R e^{i \phi} } } } \, i R e^{i \phi} \rd \phi\) | Definition of Complex Contour Integral | |||||||||||
\(\ds \) | \(=\) | \(\ds \lim \limits_{R \mathop \to 0} \int \limits_0^{2 \pi} \frac i {\map p {R e^{i \phi} } } \rd \phi\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \int \limits_0^{2 \pi} \lim \limits_{R \mathop \to 0} \frac i {\map p {R e^{i \phi} } } \rd \phi\) | Definite Integral of Limit of Uniformly Convergent Sequence of Integrable Functions | |||||||||||
\(\ds \) | \(=\) | \(\ds \int \limits_0^{2 \pi} \frac i {\map p 0} \rd \phi\) | Real Polynomial Function is Continuous | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {2 \pi i} {\map p 0}\) |
which is non-zero.
On the other hand, we have the following upper bound for the absolute value of the integral:
\(\ds \size {\oint \limits_{\gamma_R} \frac 1 {z \cdot \map p z} \rd z}\) | \(\le\) | \(\ds 2 \pi R \max \limits_{\size z \mathop = R} \paren {\frac 1 {\size {z \cdot \map p z} } }\) | Estimation Lemma for Contour Integrals | |||||||||||
\(\ds \) | \(=\) | \(\ds 2 \pi \max \limits_{\size z \mathop = R} \paren {\frac 1 {\size {\map p z} } }\) |
But this goes to zero for $R \to \infty$.
We have arrived at a contradiction.
Hence by Proof by Contradiction the assumption that $\map p z \ne 0$ for all $z \in \C$ must be wrong.
$\blacksquare$
Historical note
- 1746: A proof of this theorem was published by Jean le Rond d'Alembert. It was for some time called D'Alembert's Theorem.
However, it was later discovered that D'Alembert's proof was incorrect.
- 1799: The first correct proof of this was given by Gauss, and hence is also known as the D'Alembert-Gauss Theorem.
- 1814: The first full and rigorous proof in the field of complex numbers was published by Jean-Robert Argand.
Sources
- 1964: Murray R. Spiegel: Theory and Problems of Complex Variables ... (previous) ... (next): $1$: Complex Numbers: Polynomial Equations
- 1971: Allan Clark: Elements of Abstract Algebra ... (previous) ... (next): Introduction