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Comparing Complex and Reals

  1. Sep 14, 2014 #1
    3 + 7i < 5

    Is that a valid statement?

    Would that be taking the magnitude of 3 + 7i and comparing that to the magnitude of 5?

    Or would it be as simple as subtracting and -2 + 7i < 0
    But then what does that mean, for a complex number (-2 + 7i) to be less than zero?
     
  2. jcsd
  3. Sep 14, 2014 #2

    mfb

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    It is not a valid statement. The "<" comparison simply does not exist for complex numbers.
     
  4. Sep 14, 2014 #3
    So how would you bound a complex equation between a δ?

    Because when you are doing ε δ proofs you often work with linear functions.
    And when you get a quadratic, you simplify into linear functions and bound one of them.

    for example:

    lim f(x) = x2 - 25 = 0
    x -> 5

    0 < |x - 5| < δ → |f(x) - L| < ε
    0 < |x - 5| < δ → |x2 - 25 -0| < ε
    0 < |x - 5| < δ → |x2 - 25| < ε
    0 < |x - 5| < δ → |x - 5||x + 5| < ε
    0 < |x - 5| < δ → |g(x)||h(x)| < ε

    set δ to 1

    5 - δ < x < δ + 5
    5 - 1 < x < 1 + 5
    4 < x < 6

    upper and lower bounds, apply it to the complementary equation: h(x) = x + 5
    h(4) = 4 + 5 = 9
    h(6) = 6 + 5 = 11

    Because it is linear (x + 5), quick observation tells us it is increasing along the interval 4 < x < 6
    meaning no y-value will ever be greater than 11 along this interval for this equation, hence

    0 < |x - 5| < δ → |x - 5||x + 5| < |x - 5| 11 < ε

    and the proof can be shown

    My question is: Lets say you solve the quadratic and get imaginary roots, imaginary equations. Lets say that h(x) is a function with imaginaries within. How would I bound that function by a δ of 1, or by anything for that matter, to make sure it is not increasing without bounds?
     
  5. Sep 14, 2014 #4

    mfb

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    All those comparisons are done in the real numbers. The magnitude of a complex number is always real, and all the other numbers used there are real as well.
     
  6. Sep 14, 2014 #5

    WWGD

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    But to nitpick, there is no ordering that respects the field properties ; there are always orderings to be defined.
     
  7. Sep 14, 2014 #6
    A couple of ways.
    if you are defining a boundary of complex numbers of the form x+iy you can use functions

    You could have 2 functions y1 (x) and y2 (x)
    If the graph of y2 is higher than y1 over the entire boundary x=a to x=b, then the following inequalities would work:
    X <y2 (x)
    AND
    x> y1 (x)
    AND

    b> x> a

    This is because complex numbers form a plane with x y coordinates
     
  8. Sep 14, 2014 #7

    mfb

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    Yes you can define orderings, but there is no standard ordering, so just using "<" without an explicit non-standard definition is not meaningful.
     
  9. Sep 14, 2014 #8
    So you cannot bound an equation with imaginary parts? What you are saying is that to do that wouldn't even make sense.
     
  10. Sep 15, 2014 #9

    WWGD

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    You can always defined an order (<,S)on any set S ; a partial order is a relationship on the set that is
    i)Reflexive
    ii)Antisymmetric
    iii) Transitive

    If any two elements z,w are comparable, i.e., if you can always decide whether z<w or w<z , then the order is
    a total order; otherwise it is a partial order. Examples of partial order, maybe the most natural one, is that
    defined on the collection of subsets by inclusion. Maybe the best example of a total order is that of the Reals
    with a>b => a-b >0 .


    What you cannot always do is to have an order that is "preserved" by, or respects the field properties.
    http://en.wikipedia.org/wiki/Ordered_field

    Now, the Reals under the standard < , i.e., a<b iff (Def.) b-a>0 are an ordered field. Notice the problem comes
    from the imaginary numbers (assuming the Real part) : if i>0 , then i.i =-1 . Now you can then assume in
    your theory that -1>0 . But then (-1)(-1)=1 >0 , so 1>0 and -1>0 . But then 1+(-1)=0 >0

    Check out the linked page to see how it is not possible to turn the usual Complex numbers; usual sum and product, into an ordered field.

    Basically , a<b can mean many different things depending on the context.
     
    Last edited: Sep 15, 2014
  11. Sep 15, 2014 #10

    mfb

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    Those typical bounds are bounds for the magnitude of the complex number. The magnitude is a real value, and sure, this can have bounds.
     
  12. Sep 15, 2014 #11

    Curious3141

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    Not only is the magnitude real, but it is a non-negative real. Just wanted to add that.
     
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