# 1.7 The complex numbers  (Page 3/3)

 Page 3 / 3

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

$|z|=\sqrt{{x}^{2}+{y}^{2}}.$

We define the distance $d\left(z,w\right)$ between two complex numbers $z$ and $w$ by

$d\left(z,w\right)=|z-w|.$

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

${B}_{r}\left(c\right)=\left\{z\in C:|z-c|

The closed disk of radius $r$ around $c$ is denoted by ${\overline{B}}_{r}\left(c\right)$ and is defined by

${\overline{B}}_{r}\left(c\right)=\left\{z\in C:|z-c|\le r\right\}.$

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

${B}_{r}^{\text{'}}\left(c\right)=\left\{z:0<|z-c|

and

${\overline{B}}_{r}^{\text{'}}\left(c\right)=\left\{z:0<|z-c\le r\right\}.$

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| 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 $|z-w|\le r.$

1. Prove that the absolute value of a complex number $z$ is a nonnegative real number. Show in addition that ${|z|}^{2}=z\overline{z}.$
2. Let $x$ be a real number. Show that the absolute value of $x$ is the same whether we think of $x$ as a real number or as a complex number.
3. Prove that $max\left(|\Re \left(z\right)|,|\Im \left(z\right)|\right)\le |z|\le |\Re \left(z\right)|+|\Im \left(z\right)|.$ Note that this just amounts to verifying that
$max\left(|x|,|y|\right)\le \sqrt{{x}^{2}+{y}^{2}}\le |x|+|y|$
for any two real numbers $x$ and $y.$
4. For any complex numbers $z$ and $w,$ show that $\overline{z+w}=\overline{z}+\overline{w},$ and that $\overline{\overline{z}}=z.$
5. Show that $z+\overline{z}=2\Re \left(z\right)$ and $z-\overline{z}=2i\Im \left(z\right).$
6. If $z=a+bi$ and $w={a}^{\text{'}}+{b}^{\text{'}}i,$ prove that $|zw|=|z||w|.$ HINT: Just compute $|\left(a+bi\right)\left({a}^{\text{'}}+{b}^{\text{'}}i\right){|}^{2}.$

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.

## Triangle inequality

If $z$ and ${z}^{\text{'}}$ are two complex numbers, then

$|z+{z}^{\text{'}}|\le |z|+|{z}^{\text{'}}|$

and

$|z-{z}^{\text{'}}|\ge ||z|-|{z}^{\text{'}}||.$

We use the results contained in [link] .

$\begin{array}{ccc}\hfill |z+{z}^{\text{'}}{|}^{2}& =& \left(z+{z}^{\text{'}}\right)\overline{\left(z+{z}^{\text{'}}\right)}\hfill \\ & =& \left(z+{z}^{\text{'}}\right)\left(\overline{z}+\overline{{z}^{\text{'}}}\right)\hfill \\ & =& z\overline{z}+{z}^{\text{'}}\overline{z}+z\overline{{z}^{\text{'}}}+{z}^{\text{'}}\overline{{z}^{\text{'}}}\hfill \\ & =& {|z|}^{2}+{z}^{\text{'}}\overline{z}+\overline{{z}^{\text{'}}\overline{z}}+{|{z}^{\text{'}}|}^{2}\hfill \\ & =& {|z|}^{2}+2\Re \left({z}^{\text{'}}\overline{z}\right)+{|{z}^{\text{'}}|}^{2}\hfill \\ & \le & {|z|}^{2}+2|\Re \left({z}^{\text{'}}\overline{z}\right)|+|{z}^{\text{'}}{|}^{2}\hfill \\ & \le & {|z|}^{2}+2|{z}^{\text{'}}\overline{z}|+|{z}^{\text{'}}{|}^{2}\hfill \\ & =& {|z|}^{2}+2|{z}^{\text{'}}||z|+|{z}^{\text{'}}{|}^{2}\hfill \\ & =& \left(|z|+|{z}^{\text{'}}{|\right)}^{2}.\hfill \end{array}$

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:

$|z-w|=|z-v+v-w|\le |z-v|+|v-w|.$

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.

1. Prove the second assertion of the preceding theorem.
2. Prove the Triangle Inequality for the distance function. That is, prove that
$d\left(z,w\right)\le d\left(z,v\right)+d\left(v,w\right)$
for all $z,w,v\in C.$
3. Use mathematical induction to prove that
$|\sum _{i=1}^{n}{a}_{i}|\le \sum _{i=1}^{n}|{a}_{i}|.$

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 $|z|\le M$ for every $z$ in $S.$

1. 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] ..

2. Let $S$ be the unit circle in the plane, i.e., the set of all complex numbers $z=x+iy$ for which $|z|=1.$ Compute the sets ${S}_{1}$ and ${S}_{2}$ of part (a).

1. Verify that the formulas for the sum of a geometric progression and the binomial theorem ( [link] and [link] ) are valid for complex numbers $z$ and ${z}^{\text{'}}.$ HINT: Check that, as claimed, the proofs of those theorems work in any field.
2. Prove [link] for complex numbers $z$ and ${z}^{\text{'}}.$

how can chip be made from sand
is this allso about nanoscale material
Almas
are nano particles real
yeah
Joseph
Hello, if I study Physics teacher in bachelor, can I study Nanotechnology in master?
no can't
Lohitha
where is the latest information on a no technology how can I find it
William
currently
William
where we get a research paper on Nano chemistry....?
nanopartical of organic/inorganic / physical chemistry , pdf / thesis / review
Ali
what are the products of Nano chemistry?
There are lots of products of nano chemistry... Like nano coatings.....carbon fiber.. And lots of others..
learn
Even nanotechnology is pretty much all about chemistry... Its the chemistry on quantum or atomic level
learn
da
no nanotechnology is also a part of physics and maths it requires angle formulas and some pressure regarding concepts
Bhagvanji
hey
Giriraj
Preparation and Applications of Nanomaterial for Drug Delivery
revolt
da
Application of nanotechnology in medicine
has a lot of application modern world
Kamaluddeen
yes
narayan
what is variations in raman spectra for nanomaterials
ya I also want to know the raman spectra
Bhagvanji
I only see partial conversation and what's the question here!
what about nanotechnology for water purification
please someone correct me if I'm wrong but I think one can use nanoparticles, specially silver nanoparticles for water treatment.
Damian
yes that's correct
Professor
I think
Professor
Nasa has use it in the 60's, copper as water purification in the moon travel.
Alexandre
nanocopper obvius
Alexandre
what is the stm
is there industrial application of fullrenes. What is the method to prepare fullrene on large scale.?
Rafiq
industrial application...? mmm I think on the medical side as drug carrier, but you should go deeper on your research, I may be wrong
Damian
How we are making nano material?
what is a peer
What is meant by 'nano scale'?
What is STMs full form?
LITNING
scanning tunneling microscope
Sahil
how nano science is used for hydrophobicity
Santosh
Do u think that Graphene and Fullrene fiber can be used to make Air Plane body structure the lightest and strongest. Rafiq
Rafiq
what is differents between GO and RGO?
Mahi
what is simplest way to understand the applications of nano robots used to detect the cancer affected cell of human body.? How this robot is carried to required site of body cell.? what will be the carrier material and how can be detected that correct delivery of drug is done Rafiq
Rafiq
if virus is killing to make ARTIFICIAL DNA OF GRAPHENE FOR KILLED THE VIRUS .THIS IS OUR ASSUMPTION
Anam
analytical skills graphene is prepared to kill any type viruses .
Anam
Any one who tell me about Preparation and application of Nanomaterial for drug Delivery
Hafiz
what is Nano technology ?
write examples of Nano molecule?
Bob
The nanotechnology is as new science, to scale nanometric
brayan
nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
Damian
Got questions? Join the online conversation and get instant answers!