# 1.7 The complex numbers  (Page 2/3)

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$w=\frac{x}{{x}^{2}+{y}^{2}}+\frac{-y}{{x}^{2}+{y}^{2}}i.$

We then have

$\begin{array}{ccc}\hfill z×w& =& \left(x+yi\right)×\left(\frac{x}{{x}^{2}+{y}^{2}}+\frac{-y}{{x}^{2}+{y}^{2}}i\hfill \\ & =& \frac{{x}^{2}}{{x}^{2}+{y}^{2}}-\frac{-{y}^{2}}{{x}^{2}+{y}^{2}}+\left(x\frac{-y}{{x}^{2}+{y}^{2}}+y\frac{x}{{x}^{2}+{y}^{2}}\right)i\hfill \\ & =& \frac{{x}^{2}+{y}^{2}}{{x}^{2}+{y}^{2}}+\frac{0}{{x}^{2}+{y}^{2}}i\hfill \\ & =& 1+0i\hfill \\ & =& 1,\hfill \end{array}$

as desired.

Prove parts (1) and (2) of [link] .

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 $1+{z}^{2}?$

1. Prove that every complex number has a square root. HINT: Let $z=a+bi.$ Assume $w=x+yi$ satisfies ${w}^{2}=z,$ and just solve the two equations in two unknowns that arise.
2. Prove that every quadratic equation $a{z}^{2}+bz+c=0,$ for $a,b,$ and $c$ complex numbers, has a solution in $C.$ HINT: If $a=0,$ it is easy to find a solution. If $a\ne 0,$ we need only find a solution to the equivalent equation
${z}^{2}+\frac{b}{a}z+\frac{c}{a}=0.$
Justify the following algebraic manipulations, and then solve the equation.
$\begin{array}{ccc}\hfill {z}^{2}+\frac{b}{a}z+\frac{c}{a}& =& {z}^{2}+\frac{b}{a}z+\frac{{b}^{2}}{4{a}^{2}}-\frac{{b}^{2}}{4{a}^{2}}+\frac{c}{a}\hfill \\ & =& {\left(z+\frac{b}{2a}\right)}^{2}-\frac{{b}^{2}}{4{a}^{2}}+\frac{c}{a}.\hfill \end{array}$

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.

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