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 $\map p z$ 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


Sources

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