(related to Lemma: Complex Numbers are Two-Dimensional and the Complex Numbers \(1\) and Imaginary Unit \(i\) Form Their Basis)
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.
