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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], it holds that

[math]\displaystyle{ \int\limits_{\gamma }{f(z)dBz}=0. }[/math]

Proof: By the Cauchy-Riemann partial 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

[math]\displaystyle{ \int\limits_{\gamma }{f(z)dBz}=\int\limits_{\gamma }{\left( u+iv \right)\left( dBx+idBy \right)}=\int\limits_{z\in {{A}^{-}}}{\left( i\left( \frac{\partial Bu}{\partial Bx}-\frac{\partial Bv}{\partial By} \right)-\left( \frac{\partial Bv}{\partial Bx}+\frac{\partial Bu}{\partial By} \right) \right)dB(x,y)}=0.\square }[/math]

Fundamental theorem of algebra

For every non-constant polynomial [math]\displaystyle{ p \in {}^{(\omega)}\mathbb{C} }[/math], there exists some [math]\displaystyle{ z \in {}^{(\omega)}\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 {}^{(\omega)}\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 {}^{(\omega)}\mathbb{R}_{>0} }[/math], and hence [math]\displaystyle{ f(0) = \mathcal{O}(\text{d0}) }[/math], which is a contradiction.[math]\displaystyle{ \square }[/math]

Recommended readings

Nonstandard Mathematics