1.7 The complex numbers  (Page 2/3)

We then have

as desired.

One might think that these kinds of improvements of the real numbers will go on and on. For instance, we might next have to create and adjoin another object $j$ so that the number $i$ has a square root; i.e., so that the equation $i-z^2=0$ has a solution. Fortunately and surprisingly, this is not necessary,as we will see when we finally come to the Fundamental Theorem of Algebra in [link] .

The subset of $C$ consisting of the pairs $x+0i$ is a perfect (isomorphic) copy of the real number system $R$ . We are justifiedthen in saying that the complex number system extends the real number system, and we will say that a real number $x$ is the same as the complex number $x+0i.$ That is, real numbers are special kinds of complex numbers. The complex numbers of the form $0+yi$ are called purely imaginary numbers. Obviously, the only complex number that is both real and purely imaginary is the number $0=0+0i.$ The set $C$ can also be regarded as a 2-dimensional space, a plane, and it is also helpful to realize that the complex numbers form a 2-dimensional vector space over the fieldof real numbers.

If $z=x+yi,$ we say that the real number $x$ is the real part of $z$ and write $x=\Re \left(z\right).$ We say that the real number $y$ is the imaginary part of $z$ and write $y=\Im \left(z\right).$

If $z=x+yi$ is a complex number, define the complex conjugate $\overline{z}$ of $z$ by $\overline{z}=x-yi.$

The complex number $i$ satisfies $i^2=-1,$ showing that the negative number $-1$ has a square root in $C,$ or equivalently that the equation $1+z^2=0$ has a solution in $C.$ We have thus satisfied our initial goal of extending the real numbers. But what about other complex numbers?Do they have square roots, cube roots, $n$ th roots? What about solutions to other kinds of equations than $1+z^2?$

What about this new field $C?$ Does every complex number have a cube root, a fourth root, does every equation have a solution in $C?$ A natural instinct would be to suspect that $C$ takes care of square roots, but that it probably does not necessarily have higher order roots.However, the content of the Fundamental Theorem of Algebra, to be proved in [link] , is that every equation of the form $P\left(z\right)=0,$ where $P$ is a nonconstant polynomial, has a solution in $C.$ This immediately implies that every complex number $c$ has an $n$ th root, for any solution of the equation $z^n-c=0$ would be an $n$ th root of $c.$

The fact that the Fundamental Theorem of Algebra is true is a good indication that the field $C$ is a “good” field. But it's not perfect.

In no way can the field $C$ be made into an ordered field. That is, there exists no subset $P$ of $C$ that satisfies the two positivity axioms.