# Proof

Because we have already shown that the complex numbers form a vector space over the field of real numbers, we will now prove that the vectors complex number one $$1$$ and the imaginary unit $$i$$ are linearly independent. Let $$\alpha,\beta\in\mathbb R$$ be some real numbers and consider the equation $\alpha\cdot 1 + \beta \cdot i=0.$ We have to prove that this equation has only the trivial solution $$\alpha=\beta=0$$. For the purpose of a better understanding of the prove, we will use the notation $\alpha\cdot (1) + \beta \cdot (i)=0,$ to indicate that $$(1)$$ and $$(i)$$ are vectors and not simply numbers.

By definition, the complex vector $$1:=(1,0)$$, i.e. the complex number $$1$$ is identical with the ordered pair of real numbers $$(1,0)$$. Analogously, the complex vector $$i:=(0,1)$$, i.e. the complex number $$i$$ is identical with the ordered pair of real numbers $$(0,1)$$. Therefore, the above equation is equivalent to $\begin{array}{rcl} \alpha\cdot (1) + \beta \cdot (i)=0&\Longleftrightarrow&\beta\cdot (i)=-\alpha\cdot (1)\\ &\Longleftrightarrow&\beta\cdot (0,1)=-\alpha\cdot (1,0) \end{array}$ From the definition of the scalar multiplication, it follows $(0,\beta)=(-\alpha,0)\Longleftrightarrow -\alpha=0\text{ and }\beta=0.$

Because $$-\alpha=-1\cdot \alpha$$ (i.e the product of the real numbers $$-1$$ and $$\alpha$$) and because the product of two real numbers is only zero, if at least one of them is zero, it follows $$\alpha=0$$.

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### References

#### Bibliography

1. Forster Otto: "Analysis 1, Differential- und Integralrechnung einer VerĂ¤nderlichen", Vieweg Studium, 1983
2. Wille, D; Holz, M: "Repetitorium der Linearen Algebra", Binomi Verlag, 1994
3. Timmann, Steffen: "Repetitorium der Funktionentheorie", Binomi-Verlag, 2003