# Fundamental theorem of algebra

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