# Proof

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: $\begin{array}{ccll} (x\cdot y)\cdot z&=&[(a,b)\cdot (c,d)]\cdot (e,f)&\text{by definition of complex numbers}\\ &=&(ac-bd,ad+bc)\cdot (e,f)&\text{by definition of multiplication of complex numbers}\\ &=&((ac-bd) e-(ad+bc) f,~(ac-bd) f+(ad+bc) e)&\text{by definition of multiplication of complex numbers}\\ &=&((ace-bde)-(adf+bcf),~(acf-bdf)+(ade+bce))&\text{by distributivity law of real numbers}\\ &=&(ace-bde-adf-bcf,~acf-bdf+ade+bce)&\text{by distributivity law of real numbers and associativity of addition}\\ &=&(ace-adf-bcf-bde,~acf+ade+bce-bdf)&\text{by commutativity of addition of real numbers}\\ &=&((ace-adf)-(bcf+bde),~(acf+ade)+(bce-bdf))&\text{by distributivity law of real numbers}\\ &=&(a(ce-df)-b(cf+de),~a(cf+de)+b(ce-df))&\text{by distributivity law of real numbers}\\ &=&(a,b)\cdot(ce-df,~cf+de)&\text{by definition of multiplication of complex numbers}\\ &=&(a,b)\cdot[(c,d)\cdot(e,f)]&\text{by definition of multiplication of complex numbers}\\ &=&x\cdot (y\cdot z)&\text{by definition of complex numbers} \end{array}$

Github: ### References

#### Bibliography

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