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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 + y i is in C , we define the absolute value of z by

| z | = x 2 + y 2 .

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

d ( z , w ) = | z - w | .

If c C and r > 0 , we define the open disk of radius r around c , and denote it by B r ( c ) , by

B r ( c ) = { z C : | z - c | < r } .

The closed disk of radius r around c is denoted by B ¯ r ( c ) and is defined by

B ¯ r ( c ) = { z C : | z - c | r } .

We also define open and closed punctured disks B r ' ( c ) and B ¯ r ' ( c ) around c by

B r ' ( c ) = { z : 0 < | z - c | < r }

and

B ¯ r ' ( c ) = { z : 0 < | z - c r } .

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 ( S ) , to be the set of all z such that there exists a w S for which | z - w | < r . That is, N r ( S ) 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 N ¯ r ( S ) , to be the set of all z C for which there exists a w S such that | z - w | r .

  1. Prove that the absolute value of a complex number z is a nonnegative real number. Show in addition that | z | 2 = z 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 ( | ( z ) | , | ( z ) | ) | z | | ( z ) | + | ( z ) | . Note that this just amounts to verifying that
    max ( | x | , | y | ) x 2 + y 2 | x | + | y |
    for any two real numbers x and y .
  4. For any complex numbers z and w , show that z + w ¯ = z ¯ + w ¯ , and that z ¯ ¯ = z .
  5. Show that z + z ¯ = 2 ( z ) and z - z ¯ = 2 i ( z ) .
  6. If z = a + b i and w = a ' + b ' i , prove that | z w | = | z | | w | . HINT: Just compute | ( a + b i ) ( a ' + b ' i ) | 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 ' are two complex numbers, then

| z + z ' | | z | + | z ' |

and

| z - z ' | | | z | - | z ' | | .

We use the results contained in [link] .

| z + z ' | 2 = ( z + z ' ) ( z + z ' ) ¯ = ( z + z ' ) ( z ¯ + z ' ¯ ) = z z ¯ + z ' z ¯ + z z ' ¯ + z ' z ' ¯ = | z | 2 + z ' z ¯ + z ' z ¯ ¯ + | z ' | 2 = | z | 2 + 2 ( z ' z ¯ ) + | z ' | 2 | z | 2 + 2 | ( z ' z ¯ ) | + | z ' | 2 | z | 2 + 2 | z ' z ¯ | + | z ' | 2 = | z | 2 + 2 | z ' | | z | + | z ' | 2 = ( | z | + | z ' | ) 2 .

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 | | 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 ( z , w ) d ( z , v ) + d ( v , w )
    for all z , w , v C .
  3. Use mathematical induction to prove that
    | i = 1 n a i | 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 | 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 + i y 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 ' . HINT: Check that, as claimed, the proofs of those theorems work in any field.
  2. Prove [link] for complex numbers z and z ' .

Questions & Answers

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Eke Reply
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Almas
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Missy Reply
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Joseph
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learn
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Google
da
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Preparation and Applications of Nanomaterial for Drug Delivery
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da
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ya I also want to know the raman spectra
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what about nanotechnology for water purification
RAW Reply
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
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Alexandre
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Brian Reply
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?
LITNING Reply
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LITNING Reply
What is meant by 'nano scale'?
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LITNING
scanning tunneling microscope
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Rafiq
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Rafiq
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Any one who tell me about Preparation and application of Nanomaterial for drug Delivery
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Bob Reply
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Bob
The nanotechnology is as new science, to scale nanometric
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Damian
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Source:  OpenStax, Analysis of functions of a single variable. OpenStax CNX. Dec 11, 2010 Download for free at http://cnx.org/content/col11249/1.1
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