# Proof

By the definition of complex numbers, a complex number $$x\in\mathbb C$$ can be represented by a pair of real numbers $$a,b\in\mathbb R$$: $x:=(a,b).$

Let $$x\neq 0$$, i.e. let $$x$$ not be the complex zero. Therefore, at least one of the real numbers $$a,b$$ does not equal the real number zero $$0_{\in\mathbb R}$$. According to the rules of calculation with inequalities of real numbers, $$a^2+b^2$$ is a positive real number, and the general power. $(a^2+b^2)^{-1}=\frac 1{a^2+b^2}\quad\quad ( * )$ exists. Also, for $$b\in\mathbb R$$, the inverse number $$-b\in\mathbb R$$ with respect to addition exists. It is now obvious that we can use the real numbers $$a$$, $$-b$$ and $$( * )$$ to define a new complex number

$y:=\left(\frac{a}{a^2+b^2},\frac{-b}{a^2+b^2}\right).$

We will show that $$x^{-1}=y$$, i.e. that $$y$$ is an inverse with respect to the multiplication of complex numbers of the complex number $$x$$. We will do so applying the following mathematical definitions and concepts: * rules of multiplying positive and negative real numbers, * existence of inverse real numbers with respect to multiplication, * commutativity of multiplying real numbers , * existence of inverse real numbers with respect to addition, and * existence of the complex number one. $\begin{array}{rcll} x \cdot y&=&(a,b)\cdot\left(\frac{a}{a^2+b^2},\frac{-b}{a^2+b^2}\right)&\text{by definition of complex numbers}\\ &=&\left(\frac{a\cdot a-b\cdot (-b)}{a^2+b^2},\frac{a\cdot (-b)+b\cdot a}{a^2+b^2}\right)&\text{by definition of multiplication of complex numbers}\\ &=&\left(\frac{a^2+b^2}{a^2+b^2},\frac{-ab+ba}{a^2+b^2}\right)&\text{by the rules of multiplying positive and negative real numbers}\\ &=&(1_{\in\mathbb R},~ \frac{-ab+ba}{a^2+b^2})&\text{by existence of inverse real numbers with respect to multiplication}\\ &=&(1_{\in\mathbb R},~ \frac{-ba+ba}{a^2+b^2})&\text{by commutativity of multiplying real numbers}\\ &=&(1_{\in\mathbb R},~ \frac{0_{\in\mathbb R}}{a^2+b^2})&\text{by existence of inverse real numbers with respect to addition}\\ &=&(1_{\in\mathbb R},~ 0_{\in\mathbb R})&\text{multiplication of a real number by zero}\\ &=&1&\text{by the definition of complex one} \end{array}$ This completes the proof.

Github: ### References

#### Bibliography

1. Forster Otto: "Analysis 1, Differential- und Integralrechnung einer Veränderlichen", Vieweg Studium, 1983