why is $i^2=-1$?

Question

Question: Why is $i^2=-1$?

Comments

Many of us are taught that it is the axiomatic definition of complex algebra, and indeed, many textbooks start from this idea.

One book (Priestley, Intro to Complex Analysis) defines the result of complex multiplication, and then takes the identity $i^2=-1$ as a consequence of that definition.

Other discussions, at a level above my capability, suggest that the set of complex numbers can't be a field unless $i^2=-1$. In which case, is $i^2=-1$ the only option for the complex numbers to be a field? And historically, it seems like this is a retrospective justification.

Answer 1

The historical motivation for defining and accepting the existence of some algebraic structure called "the field of complex numbers" is that we want a field $\mathbb{F}$ with the properties that

1. $\mathbb{R}$ is a subfield of $\mathbb{F}$.
2. $\mathbb{F}$ is algebraically closed.

The latter property is the important property that results in $i^2=-1$. So, what does it mean for a field to be algebraically closed?

Algebraically Closed Field

There are many different ways to characterize an algebraically closed field, but my preferred characterization is the following. A field $\mathbb{F}$ is called algebraically closed if, for every degree $n$ polynomial $c_nX^n+c_{n-1}X^{n-1}+\cdots+c_1X^1+c_0$ with $c_0,c_1,...,c_n\in\mathbb{F}$, there exists $A,f_0,f_1,...,f_n\in\mathbb{F}$ such that $$c_nX^n+c_{n-1}X^{n-1}+\cdots+c_1X^1+c_0=A(X-f_0)(X-f_1)\cdots(X-f_n)$$ This is a desirable property for a field to have, because it means we can take the $n$th roots of any quantity in $\mathbb{F}$, and being able to do this polynomial factorization is extremely useful in the study of rational functions. In the case of the field $\mathbb{R}$, we are almost able to do this, but not quite. You will notice that every polynomial with coefficients in $\mathbb{R}$ can be factorized into polynomials of degree $1$ and quadratic polynomials. However, some quadratic polynomials are what we would call irreducible, which roughly means that they cannot be factored further into polynomials of degree $1$. Why is this the case? Because, notice that $$AX^2+BX+C=A(X^2+2PX+Q)$$ with $B=2AP$ and $Q=AC$, and notice that $$X^2+2PX+Q=X^2+2PX+P^2+Q-P^2=(X+P)^2-(P^2-Q)$$ Now, suppose for a second that there exists some $R\in\mathbb{R}$ such that $R^2=P^2-Q$. Then $$(X+P)^2-(P^2-Q)=(X+P)^2-R^2=(X+P-R)(X+P+R)$$ and the factorization is complete. But, remember what we assumed: we assumed the existence of $R$ to begin with. But let $P=0$ and $Q=1$. Then $R^2=P^2-Q=0^2-1=-1$. But we know for a fact that there exists no $R\in\mathbb{R}$ such that $R^2=-1$. This is a property of $\mathbb{R}$ that is actually not difficult to prove. And this fact is inconvenient for many, many disciplines in theoretical mathematics, applied mathematics, and the sciences. Hence, this is why we want a field $\mathbb{F}$ with properties $1$ and $2$ that I listed at the beginning of my answer. And now we know what it we need in order for both properties to be true: we require some $R\in\mathbb{F}$ such that $R^2=-1$, with $-1$ being understood as being "the real number $-1$ imported into $\mathbb{F}$". This is why most constructions of the field of complex numbers $\mathbb{C}$ are framed in terms of the equation $i^2=-1$. Without the existence of some such $i\in\mathbb{C}$, you cannot have both properties $1$ and $2$. This is also why many texts axiomatically define the complex numbers as "the smallest algebraically closed field containing the real numbers".

Answer 2

This is an answer that answers the present question by answering an other one (first). The question uses $i$ without defining it, this is the problem in the question, the only problem. A definition comes with some background, depending on book, lecture, analytic or algebraic point of view, et caetera. The common point of all these situation is the following one: we know $\Bbb R$ and really need a bigger field, a field extension of degree two of $\Bbb R$. This extension is important for both algebraic and analytic reasons. So we construct something. And the story of mathematics tells us, this is an important construction... Let us do it.

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We assume $\Bbb R$ is known, it is a field with known algebraic (field operation, order) and analytic structure (complete). We construct a field $\Bbb C$ as follows:

• elements of $\Bbb C$ are tuples $(a,b)$ of real numbers $a,b$,
• addition is on components,
• multiplication is given by $(a,b)\cdot (a',b'):=(aa'-bb',ab'+ba')$,
• there is no order, but the modulus / norm function can be defined, the modulus of $(a,b)$ is $\sqrt{a^2+b^2}\in\Bbb R_{\ge 0}$,
• convergence of sequences is defined by convergence on components, equivalently by convergence in norm.

It turns out, that the structure defined above is indeed a field, and the operations are continuous.

This is an abstract construction. Yes. Also the previous constructions involved in the structure of $\Bbb R$ - as built from the pieces $\Bbb N$, then $\Bbb Z$, then $\Bbb Q$ (algebraic steps so far), and then from $\Bbb Q$ to $\Bbb R$ (analytic step) are abstract constructions. (Do not take $\Bbb R$ as "given"!)

Now regarding the step from $\Bbb R$ to $\Bbb C$, we have a simple algebraic construction. We claim the above algebraic structure, and show the axioms.

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A posteriori, it turns out that we cannot work properly (in notation) with pairs $a,b$. So we need a better notation. Since there is a natural map $\varphi:\Bbb R\to \Bbb C$, mapping $a\in\Bbb R$ to the tuple $(a,0)\in\Bbb C$, it is natural to use the $\Bbb R$-algebra structure, and the algebra structure notation (omitting the structural map $\varphi$). Then with the definition $i=(0,1)$ we have $(a,b)=a+bi$. And of course $$i^2=(0,1)\cdot(0,1)=(0-1,0+0)=(-1,0)=-1\ .$$ And this is indeed the reason why $i^2=-1$. There is no need for an other reason.

Answer 3

In the same way that negative numbers are introduced to solve equations like $x+5=0$, complex numbers are introduced to solve equations like $x^2=-100$.

It turns out that we can get a field that includes all such solutions (allowing us to factor any degree-$n$ polynomial into a product of $n$ linear factors) by introducing just one "new" number $j$ whose square is negative and taking linear combinations $a+bj$ with $a,b$ real.

We don't need $j^2=-1$ for this to work. You could pick $j$ so that $j^2=-5$ and do complex arithmetic with representations of the form $a+bj$. Still, our field would include two solutions to the equation $x^2=-1$, and these solutions are what we call $i$ and $-i$. Unsurprisingly, this $i$ has convenient properties and is natural to think of as "the fundamental complex number".