1.7 The complex numbers  (Page 11/3)

It is useful to build from the real numbers another number system called the complex numbers. Although the real numbers $R$ have many of the properties we expect, i.e., every positive number has a positive square root,every number has a cube root, and so on, there are somewhat less prominent properties that $R$ fails to possess. For instance, negative numbers do not have square roots.This is actually a property that is missing in any ordered field, since every square is positive in an ordered field.See part (e) of [link] . One way of describing this shortcoming on the part of the real numbers is tonote that the equation $1+x^2=0$ has no solution in the real numbers. Any solution would have to be a number whose square is $-1,$ and no real number has that property. As an initial extension of the set of real numbers,why not build a number system in which this equation has a solution?

We faced a similar kind of problem earlier on. In the set $N$ there is no element $j$ such that $j+n=n$ for all $n\in N.$ That is, there was no element like 0 in the natural numbers. The solution to the problem in that case was simply to “create” something called zero, and just adjoin it to our set $N.$ The same kind of solution exists for us now. Let us invent an additional number, this time denoted by $i,$ which has the property that its square $i^2$ is $-1.$ Because the square of any nonzero real number is positive, this new number $i$ was traditionally referred to as an “imaginary” number. We simply adjoin this number to the set $R,$ and we will then have a number whose square is negative, i.e., $-1.$ Of course, we will require that our new number system should still be a field; we don't want to give up ourbasic algebraic operations. There are several implications of this requirement:First of all, if $y$ is any real number, then we must also adjoin to $R$ the number $y\times i\equiv yi,$ for our new number system should be closed under multiplication. Of course the square of $iy$ will equal $i^2{y}^2=-y^2,$ and therefore this new number $iy$ must also be imaginary, i.e., not a real number. Secondly, if $x$ and $y$ are any two real numbers, we must have in our new system a number called $x+yi,$ because our new system should be closed under addition.

Let $i$ denote an object whose square $i^2=-1.$ Let $C$ be the set of all objects that can be represented in the form $z=x+yi,$ where both $x$ and $y$ are real numbers.

Define two operations $+$ and $\times$ on $C$ as follows:

$$(x+yi)+(x^{\text{'}}+y^{\text{'}}i)=x+x^{\text{'}}+(y+y^{\text{'}})i,$$

and

$$(x+iy)(x^{\text{'}}+i{y}^{\text{'}})=x{x}^{\text{'}}+xi{y}^{\text{'}}+iy{x}^{\text{'}}+iyi{y}^{\text{'}}=x{x}^{\text{'}}-y{y}^{\text{'}}+(x{y}^{\text{'}}+y{x}^{\text{'}})i.$$

1. The two operations $+$ and $\times$ defined above are commutative and associative, and multiplication is distributive over addition.
2. Each operation has an identity: $(0+0i)$ is the identity for addition, and $(1+0i)$ is the identity for multiplication.
3. The set $C$ with these operations is a field.

We leave the proofs of Parts (1) and (2) to the following exercise. To see that $C$ is a field, we need to verify one final condition, and that is to show that if $z=x+yi\ne 0=0+0i,$ then there exists a $w=u+vi$ such that $z\times w=1=1+0i.$ Thus, suppose $z=x+yi\ne 0.$ Then at least one of the two real numbers $x$ and $y$ must be nonzero, so that $x^2+y^2>0.$ Define a complex number $w$ by