Why do we have $i^2 = -1$

Question

I wonder what is the connection between the square root of negative numbers and the complex plane.

Could you point me in the right direction?

Thanks a lot!

Answer 1

With the construction of $\mathbf C$ as the set $\mathbf R\times\mathbf R$, endowed with an addition and a multiplication defined as follows: \begin{cases}\begin{aligned} (a,b)+(a',b')&=(a+a',b+b'),\\ (a,b)(a',b')&=(aa'-bb',ab'+a'b), \end{aligned}\end{cases} one proves that $\mathbf C$ is a field and identify the real numbers as a subfield of $\mathbf C$ via the field homomorphism $\begin{aligned}[t] i:\mathbf R&\longrightarrow\mathbf C \\ x&\longmapsto (x,0) \end{aligned} $ This $i(1)=(1,0)$. Now by construction, $\mathbf C$ is an $\mathbf R$-vector space of dimension $2$, and its canonical basis is $\{(1,0),(0,1)\}$. Every element is $\mathbf C$ can indeed be written as $\;(x,y)=x(1,0)+y(0,1)$.

One identifies $(1,0)$ with the real number $1$ (it is indeed the unit element for the multiplication in $\mathbf C$), and the element $(0,1)$ is denoted $i$. It is easily checked that $$(0,1)(0,1)=(-1,0),$$ and with the above conventions/denotations, this can be written as $\;i^2=-1$.