How is $i$ defined, really? And why is $i \neq -i$?

Question

$i$ is generally defined as $\sqrt{-1}$ which is ambiguous because $\sqrt{x}$ is defined as the positive number whose square is $x$; however $i$ can't be positive since it isn't real.

Ok, so what if we defined $i$ as the number whose square is $-1$? This wouldn't make sense either because it would imply either of the following statements:

• $i = -i$
• $(-i)^2 \neq -1$

since the number whose square is $-1$ implies unicity.

This seems pretty basic point but none of the textbooks I've seen give more rigorous definitions than $i = \sqrt{-1}$.

Answer 1

If you take $\mathbb{R}$ as defined you can define the complex numbers in many ways. One would be

$\mathbb{C}:=\mathbb{R}^2$ \begin{align*} +:\mathbb{C}^2&\to\mathbb{C}\\ ((a,b),(c,d))&\mapsto x+y:=(a+c,b+d)\\ \cdot:\mathbb{C}^2&\to\mathbb{C}\\ ((a,b),(c,d))&\mapsto x\cdot y:=(ac-bd,ad+bc) \end{align*} In this case $\mathrm{i}=(0,1)$.

Answer 2

$i$ can be defined as one of the two solutions of $x^2 = -1$.

The other is defined as $-i$.

As you stated, stating that you're taking the "positive" root as $i$ makes little sense here. But it doesn't matter "which one" you pick to be $i$ (and which is left to be $-i$). Nothing at all changes when you're working in the complex plane.

Answer 3

Still not very rigorous, but slightly better definition of $i$ is a new number, such that $i^2=-1$.

This does not tell that $i$ is the only number, satisfying that equation. $-i$ will always do, too. From some higher-level algebraic point of view this is connected to the fact that complex conjugation $f(x+iy)=x-iy$ is an automorphism of the filed of complex numbers $\mathbb{C}$. This means that $i$ and $-i$ are indistinguishable considering their properties as complex numbers, and we should simply pick one of them to call $i$.

Answer 4

Here's a construction of a set that has all the properties that the complex numbers do. We define $$ U(a, b) = \begin{bmatrix} a & -b \\ b & a \end{bmatrix} $$ for $a, b \in \mathbb R$, and let $$C = \{U(a, b) \mid a, b \in \mathbb R \}.$$

Then we can define addition in $C$ by matrix addition, and multiplication in $C$ by matrix multiplication. Additive and multiplicative inverses exist, and the distributive and associative laws hole. $U(0,0)$ is an additive identity. $U(1, 0)$ is a multiplicative identity. The reals are a natural subgroup of $C$, with the map $r \mapsto U(r, 0)$ being an injection from $\mathbb R \to C$.

I claim that $C$ "looks exactly like" the complex numbers.

I define $i_C$ to be $U(0, 1)$. That's the thing I'll say corresponds to $i$ in the complex numbers. That's clear and unambiguous, I hope. :) Now my correspondence is $$ a + bi \mapsto a U(1, 0) + b U(0, 1) = a U(1, 0) + b i_C. $$ It's not hard to show that this is an isomorphism of fields (if you believe that the usual "definition" of $\mathbb C$ makes sense!), and now you know what $i$ is...it's just $i_C$.

If you don't beleive that the ordinary definition makes sense, then you can replace it with this one, and just say that $\mathbb C$ is $C$, because they have all the same properties. :)

Answer 5

The most common way to see the complex numbers is as pairs $(x,y) $ of reals with the operations $$ (a,b)+(c,d)=(a+c,b+d),\ \ \ \ (a,b)(c,d)=(ac-bd,ad+bc). $$ Then you identify the reals with the numbers $(a,0) $ and define $i=(0,1) $. In this case, $-i=(0,-1) $.

Answer 6

Simple version: $i = i$

$i^2 = -1$

$i^3 = -i$

$i^4 = 1$

Complex version: Essentially, complex numbers are two dimensional.

Imagine an XY plane not where "x" is one set of numbers, "y" is another set of numbers, and f(x) is the function that turns a first "x" number into a second "y" number, but instead where both "x" and "y" are used to describe the 2-dimensional coordinates of a single number: complex numbers take the form of a+bi (or bi+a, both are exactly the same), and the XY coordinates of each number is (a, b):

1 becomes 1 + 0i becomes (1, 0)

i becomes 0 + 1i becomes (0, 1)

-1 becomes -1 + 0i becomes (-1, 0)

-i becomes 0 - 1i becomes (0, -1)

Multiplying a number by 1 means that it stays the same

Multiplying a number by i [for example: $(2+i)*i = 2i-1$] means that it rotates 90 degrees counter-clockwise [the coordinates go from (2, 1) to (-1, 2)]

Multiplying a number by -1 [for example: $(2+i)*-1 = -2-i$] means that it rotates 180 degrees-counter clockwise ([the coordinates go from (2, 1) to (-2, -1)])

Multiplying a number by -i [for example: $(2+i)*-i = -2i+1$] means that it rotates 270 degrees counter-clockwise ([the coordinates go from (2, 1) to (1, -2)])

The coordinates (0, 1) and (0, -1), which correspond to i and -i respectively, are not equal to each other, but starting at (0, 1) and rotating 90 degrees counter-clockwise (starting with i and multiplying by i) gives the same number/coordinates as starting at (0, -1) and rotating 270 degrees counter-clockwise (starting with -i and multiplying by -1). Both give you the number -1 with the coordinates (-1, 0)

Answer 7

The complex numbers should be defined as the field with underlying set $\mathbb{R}\times \mathbb{R}$, with addition defined by $(a,b)+(c,d)=(a+c,b+d)$ and multiplication defined by $(a,b)(c,d) = (ac-bd,ad+bc)$. After that one defines $i=(0,1)$ with no ambiguity, and of course it satisfies $i^2=-1$

Answer 8

$\sqrt{-1}$ is not a definition, but the codeword for a problem.

The system ${\mathbb R}$ leaves much to desire, insofar as the equation $x^2=-1$, let alone more complicated equations of this sort, have no solution in ${\mathbb R}$. This leads us to look for an extension $C\supset {\mathbb R}$ that contains an element, call it $i$, with $i^2=-1$. This $C$ should be a field. The rules of algebra then immediately imply that it contains all numbers of the form $x+iy$ with real $x$, $y$, that $x+iy=u+iv$ iff $x=u$ and $y=v$, and that $$\eqalign{(x+iy)+(u+iv)&=(x+u)+ \ i(y+v),\cr (x+iy)\cdot(u+iv)&=(xu-yv)+i(xv+yu)\ .\cr}$$ Furthermore one easily verifies that $$(x+iy)\cdot\left({x\over x^2+y^2}+i\>{-y\over x^2+y^2}\right)=1\qquad(x+iy\ne0)\ .$$ This shows that the numbers $x+iy$ by themselves already form a field, which is then obviously the smallest field containing a "square root of $-1\>$", and is designed by ${\mathbb C}$.

It is a great miracle that in this field ${\mathbb C}$ not only the equation $x^2+1=0$ we started with has a solution, but any polynomial equation $p(x)=0$ of degree $\geq1$.