Fundamental Theorem of Algebra

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Theorem

Every non-constant polynomial with coefficients in $\C$ has a root in $\C$.


Proof using Algebraic Topology

Let $p \left({z}\right)$ be a polynomial in $\C$:

$p \left({z}\right) = z^m + a_1 z^{m-1} + \cdots + a_m$

where not all of $a_1, \ldots, a_m$ are zero.

Define a homotopy:

$p_t \left({z}\right) = t p \left({z}\right) + \left({1-t}\right) z^m$

Then:

$\displaystyle \frac{p_t \left({z}\right)}{z^m} = 1 + t \left({a_1 \frac 1 z + \cdots + a_m \frac 1 {z^m}}\right)$

The terms in the parenthesis go to $0$ as $z \to \infty$.


Therefore, there is an $r \in \R_{>0}$ such that:

$\forall z \in \C: \left|{z}\right| = r: \forall t \in \left[{0 \,.\,.\, 1}\right]: p_t \left({z}\right) \ne 0$

Hence the homotopy:

$\displaystyle \frac{p_t}{ \left|{p_t}\right|}:S \to \Bbb S^1$

is well-defined for all $t$.


This shows that for any complex polynomial $p \left({z}\right)$ of order $m$, there is a circle $S$ of sufficiently large radius in $\C$ such that $\dfrac{p \left({z}\right)}{\left|{p \left({z}\right)}\right|}$ and $\dfrac{z^m}{\left|{z^m}\right|}$ are freely homotopic maps $S \to \Bbb S^1$.

Hence $\dfrac{p \left({z}\right)}{\left|{p \left({z}\right)}\right|}$ must have the same degree of $\left({z / r}\right)^m$, which is $m$.

When $m > 0$, that is $p$ is non-constant, this result and the Extendability Theorem for Intersection Numbers imply $\dfrac{p \left({z}\right)}{\left|{p \left({z}\right)}\right|}$ does not extend to the disk $\operatorname{int} \left({S}\right)$, implying $p \left({z}\right) = 0$ for some $z \in \operatorname{int} \left({S}\right)$.


$\blacksquare$


Proof using Liouville's Theorem

Let $p: \C \to \C$ be a complex polynomial with $p \left({z}\right) \ne 0$ for all $z \in \C$.

Then $p$ extends to a continuous transformation of the Riemann sphere:

$\hat \C = \C \cup \left\{{\infty}\right\}$

This extension also has no zeroes.



By Riemann Sphere is Compact, there is some $\epsilon \in \R_{>0}$ such that:

$\forall z \in \C: \left|{p \left({z}\right)}\right| \ge \epsilon$


Now consider the holomorphic function $g: \C \to \C$ defined by:

$g \left({z}\right) := \dfrac 1 {p \left({z}\right)}$

We have:

$\forall z \in \C: \left|{g \left({z}\right)}\right| \le \dfrac 1 \epsilon$

By Liouville's Theorem, $g$ is constant.

Hence $p$ is also constant, as claimed.

$\blacksquare$


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:

$\displaystyle \oint \limits_{\gamma_R} \frac 1 {z \cdot \map p z} \rd z$

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 the Residue Theorem), using the parameterization $z = R e^{i \phi}$ of $\gamma_R$:

\(\displaystyle \lim \limits_{R \mathop \to 0} \oint \limits_{\gamma_R} \frac 1 {z \cdot \map p z} \rd z\) \(=\) \(\displaystyle \frac 1 {\map p 0} \lim \limits_{R \mathop \to 0} \int \limits_0^{2 \pi} \frac 1 {R e^{i \phi} } \, i R e^{i \phi} \rd \phi\) Polynomial is Continuous and Product Rule
\(\displaystyle \) \(=\) \(\displaystyle \lim_{R \mathop \to 0} \frac 1 {\map p 0} \int_0^{2 \pi} i \rd \phi\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {2 \pi i} {\map p 0}\)

which is non-zero.


On the other hand, we have the following upper bound for the absolute value of the integral:

\(\displaystyle \size {\oint \limits_{\gamma_R} \frac 1 {z \cdot \map p z} \rd z}\) \(\le\) \(\displaystyle 2 \pi R \max \limits_{\size z \mathop = R} \paren {\frac 1 {\size {z \cdot \map p z} } }\) Estimation Lemma
\(\displaystyle \) \(=\) \(\displaystyle 2 \pi \max \limits_{\size z \mathop = R} \paren {\frac 1 {\size {\map p z} } }\)

But this goes to zero for $R \to \infty$.

We arrive at a contradiction.

Hence the assumption that $\map p z \ne 0$ for all $z \in \C$ must be wrong.

$\blacksquare$


Historical Note

A proof of the Fundamental Theorem of Algebra was published in $1746$ 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.


The first correct proof was published by Carl Friedrich Gauss in his doctoral dissertation in $1799$:

Demonstratio nova theorematis omnem functionem algebraicam rationalem integram unius variabilis in factores reales primi vel secundi gradus resolvi posse (A new proof of the theorem that every integral rational algebraic function of one variable can be resolved into real factors of the first or second degree).

During the course of his career, he gave a total of four proofs of this theorem.


The first full and rigorous proof in the field of complex numbers was published in $1814$ by Jean-Robert Argand.


Sources