Fundamental Theorem of Algebra/Proof 3

From ProofWiki
Jump to navigation Jump to search

Theorem

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


Proof

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.

$\blacksquare$