(related to Proposition: Multiplication of Complex Numbers Is Associative)
By the definition of complex numbers, the complex numbers \(x,y,z \in \mathbb C\) are identified by prdered pairs \(x:=(a,b)\), \(y:=(c,d)\) and \(z:=(e,f)\) of some real numbers \(a,b,c,d,e,f\in\mathbb R\).
We have to show that \((x\cdot y)\cdot z=x\cdot (y\cdot z)\) is valid for all \(x,y,z\in\mathbb C\).
Multiplying the complex numbers \(x\cdot y\), respectively \(y\cdot z\), means by definition. \[\begin{array}{cl} x\cdot y:=&(a,b) \cdot (c,d)=(ac-bd,ad+bc),\\ y\cdot z:=&(c,d) \cdot (e,f)=(ce-df,cf+de), \end{array} \] with some new complex numbers \((ac-bd,ad+bc)\) and \((ce-df,cf+de)\). Because adding real numbers is commutative and associative, and due to the distributivity law of real numbers we have that: \[\begin{array}{ccll} (x\cdot y)\cdot z&=&[(a,b)\cdot (c,d)]\cdot (e,f)&\text{by definition of complex numbers}\\ &=&(ac-bd,ad+bc)\cdot (e,f)&\text{by definition of multiplication of complex numbers}\\ &=&((ac-bd) e-(ad+bc) f,~(ac-bd) f+(ad+bc) e)&\text{by definition of multiplication of complex numbers}\\ &=&((ace-bde)-(adf+bcf),~(acf-bdf)+(ade+bce))&\text{by distributivity law of real numbers}\\ &=&(ace-bde-adf-bcf,~acf-bdf+ade+bce)&\text{by distributivity law of real numbers and associativity of addition}\\ &=&(ace-adf-bcf-bde,~acf+ade+bce-bdf)&\text{by commutativity of addition of real numbers}\\ &=&((ace-adf)-(bcf+bde),~(acf+ade)+(bce-bdf))&\text{by distributivity law of real numbers}\\ &=&(a(ce-df)-b(cf+de),~a(cf+de)+b(ce-df))&\text{by distributivity law of real numbers}\\ &=&(a,b)\cdot(ce-df,~cf+de)&\text{by definition of multiplication of complex numbers}\\ &=&(a,b)\cdot[(c,d)\cdot(e,f)]&\text{by definition of multiplication of complex numbers}\\ &=&x\cdot (y\cdot z)&\text{by definition of complex numbers} \end{array} \]
