Fundamental Theorem of Algebra/Proof 5

Proof
Let $p(z)$ be an arbitrary non-constant polynomial with coefficients in $\mathbb{C}$. This means that for some $n \geq 1$ and $a_n \neq 0$,
 * $\ds p(z) = a_nz^n + a_{n - 1}z^{n - 1} + \dots + a_0.$

We have
 * $\ds p(z) = a_nz^n(1 + \frac{a_{n - 1}}{a_n}z^{-1} + \dots + \frac{1}{a_n}z^{-n}) = a_nz^n(1 + O(z^{-1})), \text{ as } |z| \to \infty.$

Hence
 * $\ds \lim_{|z| \to \infty}|p(z)| = \infty.$

Thus we can pick $R > 0$ large enough so that
 * $\ds \inf_{|z| \geq R}|p(z)| > |p(0)|.$

It follows that
 * $\ds \inf_{z \in \mathbb{C}}|p(z)| = \inf_{|z| \leq R}|p(z)|.$

Since $D_R = \{z \in \mathbb{C} : |z| \leq R\}$ is compact and $p$ is continuous, there exists $z_0 \in D_R$ such that
 * $\ds \inf_{|z| \leq R}|p(z)| = |p(z_0)|.$

Hence
 * $\ds |p(z_0)| = \inf_{z \in \mathbb{C}}|p(z)|.$

It remains to show that $p(z_0) = 0$. For contradiction, suppose
 * $\ds p(z_0) = a \neq 0.$

Write
 * $\ds p(z_0 + \zeta) = a + q(\zeta).$

Note that $q(\zeta)$ is a nonconstant polynomial in $\zeta$ satisfying $q(0) = 0$. Hence for some $k \geq 1$ and $b \neq 0$, we have $q(\zeta) = b\zeta^k + b_{k + 1}\zeta^{k + 1} + \dots + b_n\zeta^n$, that is,
 * $\ds q(\zeta) = b\zeta^k + O(\zeta^{k + 1}), \text{ as }\zeta \to 0.$

Thus, uniformly in $\omega \in S^1 = \{\omega \in \mathbb{C} : |\omega| = 1\}$,
 * $\ds p(z_0 + \varepsilon\omega) = a + b\omega^k\varepsilon^k + O(\varepsilon^{k + 1}), \text{ as } \varepsilon \searrow 0.$

Since $\left|-\frac{a/|a|}{b/|b|}\right| = 1$, by Euler's theorem, we can pick $\omega \in S^1$ such that $\omega^k = -\frac{a/|a|}{b/|b|}$. Then
 * $\ds p(z_0 + \varepsilon\omega) = a - a\frac{|b|}{|a|}\varepsilon^k + O(\varepsilon^{k + 1}) = a\left(1 - \frac{|b|}{|a|}\varepsilon^k\right) + O(\varepsilon^{k + 1}).$

Thus
 * $\ds |p(z_0 + \varepsilon\omega)| = |a|\left|1 - \frac{|b|}{|a|}\varepsilon^k\right| + O(\varepsilon^{k + 1}).$

Thus for $\varepsilon > 0$ small enough, \begin{align} &= |a| - |b|\varepsilon^k + O(\varepsilon^{k + 1}) \\ &= |a| - |b|\varepsilon^k(1 + O(\varepsilon)) \\ &< |a|. \\ \end{align} But this contradicts that $|a| = \inf_{z \in \mathbb{C}}|p(z)|$. Thus $p(z_0) = a = 0$.
 * p(z_0 + \varepsilon\omega)| &= |a|\left(1 - \frac{|b|}{|a|}\varepsilon^k\right) + O(\varepsilon^{k + 1}) \\