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Suppose $C$ were an ordered field, and write $P$ for its set of positive elements. Then, since every square in an ordered field must be in $P$ (part (e) of [link] ), we must have that $-1={i}^{2}$ must be in $P.$ But, by part (a) of [link] , we also must have that 1 is in $P,$ and this leads to a contradiction of the law of tricotomy. We can't have both 1 and $-1$ in $P.$ Therefore, $C$ is not an ordered field.
Although we may not define when one complex number is smaller than another, we can define the absolute value of a complex number and the distance between two of them.
If $z=x+yi$ is in $C$ , we define the absolute value of $z$ by
We define the distance $d(z,w)$ between two complex numbers $z$ and $w$ by
$d(z,w)=|z-w|.$
If $c\in C$ and $r>0,$ we define the open disk of radius $r$ around c , and denote it by ${B}_{r}\left(c\right),$ by
The closed disk of radius $r$ around $c$ is denoted by ${\overline{B}}_{r}\left(c\right)$ and is defined by
We also define open and closed punctured disks ${B}_{r}^{\text{'}}\left(c\right)$ and ${\overline{B}}_{r}^{\text{'}}\left(c\right)$ around $c$ by
and
These punctured disks are just like the regular disks, except that they do not contain the central point $c.$
More generally, if $S$ is any subset of $C,$ we define the open neighborhood of radius r around $S,$ denoted by ${N}_{r}\left(S\right),$ to be the set of all $z$ such that there exists a $w\in S$ for which $|z-w|<r.$ That is, ${N}_{r}\left(S\right)$ is the set of all complex numbers that are within a distance of $r$ of the set $S.$ We define the closed neighborhood of radius $r$ around $S,$ and denote it by ${\overline{N}}_{r}\left(S\right),$ to be the set of all $z\in C$ for which there exists a $w\in S$ such that $|z-w|\le r.$
The next theorem is in a true sense the most often used inequality of mathematical analysis. We have already proved the triangle inequality for the absolute value of real numbers, and the proof wasnot very difficult in that case. For complex numbers, it is not at all simple, and this should be taken as a good indication that it is a deep result.
If $z$ and ${z}^{\text{'}}$ are two complex numbers, then
and
We use the results contained in [link] .
The Triangle Inequality follows now by taking square roots.
REMARK The Triangle Inequality is often used in conjunction with what's called the “add and subtract trick.”Frequently we want to estimate the size of a quantity like $|z-w|,$ and we can often accomplish this estimation by adding and subtracting the same thing within the absolute value bars:
The point is that we have replaced the estimation problem of the possibly unknown quantity $|z-w|$ by the estimation problems of two other quantities $|z-v|$ and $|v-w|.$ It is often easier to estimate these latter two quantities, usually by an ingenious choice of $v$ of course.
It may not be necessary to point out that part (b) of the preceding exercise provides a justification for the name “triangle inequality.”Indeed, part (b) of that exercise is just the assertion that the length of one side of a triangle in the plane is less than or equalto the sum of the lengths of the other two sides. Plot the three points $z,w,$ and $v,$ and see that this interpretation is correct.
A subset $S$ of $C$ is called Bounded if there exists a real number $M$ such that $\left|z\right|\le M$ for every $z$ in $S.$
Let $S$ be a subset of $C$ . Let ${S}_{1}$ be the subset of $R$ consisting of the real parts of the complex numbers in $S,$ and let ${S}_{2}$ be the subset of $R$ consisting of the imaginary parts of the elements of $S.$ Prove that $S$ is bounded if and only if ${S}_{1}$ and ${S}_{2}$ are both bounded.
HINT: Use Part (c) of [link] ..
Let $S$ be the unit circle in the plane, i.e., the set of all complex numbers $z=x+iy$ for which $\left|z\right|=1.$ Compute the sets ${S}_{1}$ and ${S}_{2}$ of part (a).
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