Fundamental Theorem of Algebra/Proof 6
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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 $z \in \C$.
Therefore by Reciprocal of Holomorphic Function $\dfrac 1 {\map p z}$ is entire.
\(\ds 0\) | \(<\) | \(\ds \cmod {\dfrac 1 {\map p 0} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \cmod {\frac 1 {2 \pi i} \oint_{\set {w \in \C : \cmod w = R} } \frac {\rd z} {z \cdot \map p z} }\) | Cauchy's Integral Formula | |||||||||||
\(\ds \) | \(\le\) | \(\ds \frac 1 {2 \pi} \max \limits_{\size z \mathop = R} \paren {\frac 1 {\size {z \cdot \map p z} } } 2 \pi R\) | Estimation Lemma for Contour Integrals | |||||||||||
\(\ds \) | \(=\) | \(\ds \max \limits_{\size z \mathop = R} \frac 1 {\size {\map p z} }\) | ||||||||||||
\(\ds \) | \(\to\) | \(\ds 0\) | as $R \to +\infty$ |
This is a contradiction.
$\blacksquare$