# Proof

We want to show that in the complex vector space of complex numbers over the field or real numbers, the set of vectors $$B:=\{1,i\}$$ forms a basis, i.e. any complex number $$x\in\mathbb C$$ can be represented by a linear combination of the vectors:

$x=a \cdot 1+b \cdot i$

for some real numbers $$a,b\in\mathbb R$$.

We have already shown that the vectors $$B:=\{1,i\}$$ are linearly independent. It remains to be shown that $$B$$ is a generating system of $$\mathbb C$$, i.e. that the set of all possible linear combinations of the vectors in $$B$$, i.e. the linear span

$Span(B):=\{a \cdot 1+ b \cdot i~|~a,b\in\mathbb R,~1,i\in \mathbb C\}$

generates the complex numbers $$\mathbb C$$. By definition, the complex vector $$1$$ equals the ordered pair of real numbers $$(1,0)$$. Analogously, the complex vector $$i$$ is identical with the ordered pair of real numbers $$(0,1)$$. From the definition of scalar multiplication, and the addition of complex numbers it follows $a \cdot 1+ b \cdot i=a \cdot (1,0)+ b \cdot (0,1)=(a,0)+(0,b)=(a,b)$ for all real numbers $$a,b\in\mathbb R$$. Therefore, any complex number $$x=(a,b)$$ is exactly the number, which is represented by the linear combination $$a \cdot 1+ b \cdot i$$. It follows $Span(B)=\mathbb C.$

As a consequence, $$B$$ is a basis of $$\mathbb C$$. The maximum number of linearly independent vectors, which can be spanned in $$\mathbb C$$ is $$2$$ (otherwise, we would have $$Span(B)\subset \mathbb C$$ and not $$Span(B)=\mathbb C$$. Thus, $$\mathbb C$$ is two-dimensional.

Github: ### References

#### Bibliography

1. Wille, D; Holz, M: "Repetitorium der Linearen Algebra", Binomi Verlag, 1994
2. Timmann, Steffen: "Repetitorium der Funktionentheorie", Binomi-Verlag, 2003