(related to Proposition: Existence of Inverse Complex Numbers With Respect to Multiplication)
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.
