Fundamental theorem of algebra

For every non-constant polynomial [math]\displaystyle{ p \in \mathbb{C} }[/math], there exists some [math]\displaystyle{ z \in \mathbb{C} }[/math] such that [math]\displaystyle{ p(z) = 0 }[/math].

Indirect proof: By performing an affine substitution of variables, reduce to the case [math]\displaystyle{ 1/p(0) \ne \mathcal{O}(\text{d0}) }[/math]. Suppose that [math]\displaystyle{ p(z) \ne 0 }[/math] for all [math]\displaystyle{ z \in \mathbb{C} }[/math]. Since [math]\displaystyle{ f(z) := 1/p(z) }[/math] is holomorphic, it holds that [math]\displaystyle{ f(1/\text{d0}) = \mathcal{O}(\text{d0}) }[/math]. By the mean value inequality [math]\displaystyle{ |f(0)| \le {|f|}_{\gamma} }[/math] for [math]\displaystyle{ \gamma = \partial\mathbb{B}_{r}(0) }[/math] and arbitrary [math]\displaystyle{ r \in \mathbb{R}_{>0} }[/math], and hence [math]\displaystyle{ f(0) = \mathcal{O}(\text{d0}) }[/math], which is a contradiction.[math]\displaystyle{ \square }[/math]

See also

• List of mathematical symbols
• Fundamental theorem of algebra