Theorem 27.4.4 (Fundamental Theorem of Algebra).label Let $p \in \complex[z]$ be a non-constant polynomial, then there exists $z \in \complex$ such that $f(z) = 0$.
Proof. Let $p \in \complex[z]$ such that $p(z) \ne 0$ for all $z \in \complex$. Let $f = 1/p$, then $f \in H(\complex; \complex) \cap C_{0}(\complex; \complex)$, so $f$ is bounded. By Liouville’s Theorem, $f$ and thus $p$ is constant.$\square$
Post a Comment