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Math Help - bigger of two complex numbers and why?

  1. #1
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    bigger of two complex numbers and why?

    which one of these two complex numbers is big?
    2+3i and 3+2i and why?
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  2. #2
    Senior Member eumyang's Avatar
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    Define "bigger." Because I have no idea what you mean.

    Are you perhaps talking about the modulus of a complex number? The modulus for a complex number a + bi is
    |a + bi| = \sqrt{a^2 + b^2}.
    (If you were to plot the point representing a complex number on the complex plane, the modulus is also the distance between that point and the origin.)

    So plug in the numbers into the expression above to find the moduli (?) of the two complex numbers you have and you can find out which one is "bigger."
    Last edited by mr fantastic; July 4th 2010 at 03:13 PM. Reason: I moved the thread.
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  3. #3
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    Quote Originally Posted by vg284 View Post
    which one of these two complex numbers is big?
    2+3i and 3+2i and why?
    The complex numbers cannot be made into an ordered field so it makes no sense to ask which is bigger. The question should read

    Which one of these complex numbers has the biggest magnitude
    and you have been told in post #2 how to do this.

    (In fact, they both have the same magnitude so actually the question makes absolutely no sense. I suggest you point this out to your teacher).
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  4. #4
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    thanks for your response.

    my question was


    as 2<3 and 4>2 etc. Can we say that (2+3i)>(3+2i) or (2+3i)<(3+2i)?
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  5. #5
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    Quote Originally Posted by vg284 View Post
    Can we say that (2+3i)>(3+2i) or (2+3i)<(3+2i)?
    You can't either of these. You can say |2+3i| = |3+2i|
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  6. #6
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    The set of complex numbers cannot be ordered, as uemyang and pickslides said, in such a way that the usual rules for "order" are still true:
    "If a> b then a+ c> b+ c for all c"
    "If a> b then ac> bc for all c> 0"
    "For any a and b, one and only one must be true:
    (i) a= b
    (ii) a> b
    (iii)b> a".

    Suppose we had defined an order ">". Then, since i\ne 0, either i> 0 or i< 0.

    1) Suppose i> 0. Then, multiplying both sides by the positive number i, i(i)> 0(i) or -1> 0. That by itself is NOT a contradiction since this order on the complex numbers may not give the "usual" order on the real numbers. However, now we have -1(i)> 0(i) or -i> 0. Adding i to both sides, 0> i which does contradict i> 0.

    2) Suppose 0> i. Then, subtracting i from both sides 0- i> i- i or -i> 0. Multiplying both sides by the positive number -i, (-i)(-i)= i(i)> 0(-i) or -1> 0. Again, that is not necessarily a contradiction. But now, -i(-1)> -i(0) gives i> 0 which contradicts 0> i.

    Either way, we arrive at a contradiction so there is no way to order the complex numbers.
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