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Theorems of the month
Cauchy's integral theorem
Given the neighbourhood relations [math]\displaystyle{ B \subseteq {A}^{2} }[/math] and [math]\displaystyle{ D \subseteq [a, b] }[/math] for some simply connected [math]\displaystyle{ h }[/math]-set [math]\displaystyle{ A \subseteq {}^{\omega}\mathbb{C} }[/math], infinitesimal [math]\displaystyle{ h }[/math], a holomorphic function [math]\displaystyle{ f: A \rightarrow {}^{\omega}\mathbb{C} }[/math] and a closed path [math]\displaystyle{ \gamma: [a, b[\rightarrow \partial A }[/math], choosing [math]\displaystyle{ \curvearrowright B \gamma(t) = \gamma(\curvearrowright D t) }[/math] for [math]\displaystyle{ t \in [a, b[ }[/math] gives
Proof: By the Cauchy-Riemann differential equations and Green's theorem, with [math]\displaystyle{ x := \text{Re} \, z, y := \text{Im} \, z, u := \text{Re} \, f, v := \text{Im} \, f }[/math] and [math]\displaystyle{ {A}^{-} := \{z \in A : z + h + ih \in A\} }[/math], it holds that
Fundamental theorem of algebra
Every non-constant polynomial [math]\displaystyle{ p \in {}^{(\omega)}\mathbb{C} }[/math] has at least one complex root.
Indirect proof: By performing an affine substitution of variables, reduce to the case [math]\displaystyle{ \widetilde{p(0)} \ne \mathcal{O}(\iota) }[/math]. Suppose that [math]\displaystyle{ p(z) \ne 0 }[/math] for all [math]\displaystyle{ z \in {}^{(\omega)}\mathbb{C} }[/math].
Since [math]\displaystyle{ f(z) := \widetilde{p(z)} }[/math] is holomorphic, it holds that [math]\displaystyle{ f(\tilde{\iota}) = \mathcal{O}(\iota) }[/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 {}^{(\omega)}\mathbb{R}_{>0} }[/math], and hence [math]\displaystyle{ f(0) = \mathcal{O}(\iota) }[/math], which is a contradiction.[math]\displaystyle{ \square }[/math]