Why must be $i^2=-1$?

Question

These days I am explaining radicals to my 15-year-old students, and I wanted to add that square roots or roots with even index and negative radical can be solved with complex numbers. I know that n mathematics, the imaginary unit $i$ (sometimes represented by the Greek letter ($\iota$) makes it possible to extend the range of field numbers $\mathbb {R}$ to the field of complex numbers $\mathbb {C}$. The imaginary unit is characterized by being a number whose square is equal to $-1$. The powers of $i$ repeat periodically (they are cyclic with period $4$).

Is there a real reason why $$i^2=-1\,?$$

Or it is like a postulate that we must assume. Sorry for the trivial question.

Answer 1

This is simply the definition of complex numbers. But I would say that this is just one of the available extensions of the real numbers. We can denote $\mathbb{C} = \mathbb{R}[i] = \{a + bi: a, b \in \mathbb{R}\}$. Similarly:

• dual numbers: $\mathbb{R[\varepsilon]} = \{a + b\varepsilon: a, b \in \mathbb{R}\}$, where $\varepsilon^2 = 0, \varepsilon \neq 0$;
• split complex numbers: $\mathbb{R}[j] = \{a + bj: a, b \in \mathbb{R}\}$, where $j^2 = 1, j \notin \mathbb{R}$.

So why do we use complex numbers and not the other two? One sensible explanation is that the complex numbers are the only ones of the three to form a field.