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Complex numbers and negative roots

  1. Aug 10, 2015 #1
    I was wondering if scientists or mathematicians have any use for complex numbers involving negative roots of I as in i=(-1)^(1/2). but my question is more what would be (-1)^(-1/2)?
     
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  3. Aug 10, 2015 #2

    mfb

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    Complex numbers are used everywhere. Making your computer would have been much harder without complex numbers in circuit design and quantum mechanics, for example.
    Also ##\pm i##, as ##\frac{1}{i}=-i## which you can see if you multiply both sides by i.
     
  4. Aug 10, 2015 #3
    Yes, they are all over the place in math and physics. Indeed, there are entire branches of physics that would be impossible to do without them.

    $$\sqrt{-1}$$

    is represented by the letter i.

    The wikipedia page has many more details:
    https://en.wikipedia.org/wiki/Complex_number
     
  5. Aug 12, 2015 #4

    HOI

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    By the way, there are problems involved with defining "[itex]i= \sqrt{-1}[/itex]'. It can be shown that, in the complex numbers, all numbers have two square roots- so that notation is ambiguous. A better way to define the complex numbers is as pairs of real numbers, (a, b) with addition defined by (a, b)+ (c, d)= (a+ c, b+ d) and multiplication defined by (ab- cd, ad+ bc).

    It can be show that this is a field with additive identity (0, 0) and multiplicative identity (1, 0). Further, the field of real numbers can be identified with the subfield (a, 0).
    If we then represent multiplication of a real number, a, by a complex number, (b, c) with (a, 0)(b, c) as (ab,, ac), every complex number can be written in the form (a, b)= (a, 0)+ (0, b)= a(1, 0)+ b(0, 1). We have already agreed to represent the real number "1" by "(1, 0)". If we now agree to write [itex]i= (0, 1)[/itex] we have (a, b)= a+ bi.
     
  6. Aug 17, 2015 #5
    I don't understand what you mean when you say this notation is ambiguous. The radical sign means principal square root.
     
  7. Aug 17, 2015 #6

    symbolipoint

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    Better to start with [itex] i^2=-1[/itex].
     
  8. Aug 17, 2015 #7
    sorry, I dont understand the point you are making about ambiguity.
     
  9. Aug 17, 2015 #8

    symbolipoint

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    Exponentiate both sides to the [itex] \frac{1}{2}[/itex] power.
    Plus or Minus Square Root of negative 1; no more ambiguity.
     
  10. Aug 17, 2015 #9
    Could you write in greater detail: that is equally confusing. HOI said the notation is ambiguous, and I do not understand if you are also saying the notation is ambiguous. It seems to me you are implying that ##i = \pm \sqrt{-1}##. Further, I do not understand why you would not use ##x^2 = -1##, and then label the principal root ##i## such that ##x_1 = i## and ##x_2 = -i##. There does not appear to be anything ambiguous about this process to me.
     
  11. Aug 17, 2015 #10

    symbolipoint

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    Then why you say it was ambiguous? You made the correct "implication".
     
  12. Aug 17, 2015 #11
    HOI said it was ambiguous. I am asking for clarification about what makes it ambiguous to HOI or anyone who agrees that it is ambiguous.

    I do not agree that ##i = \pm \sqrt{-1}##.
     
  13. Aug 17, 2015 #12

    symbolipoint

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    I do not understand you. You also decided that i=sqrt(-1) was ambiguous, and then you found the correct meaning from i^2. The meaning of i is either ambiguous or it is not.

    You are disagreeing with a very logically derived concept or fact.
     
  14. Aug 17, 2015 #13

    symbolipoint

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    thelema418,
    Maybe I see the trouble. You MUST understand signed numbers.
    Example, 5, the real number, integer number, 5.

    (5)(5)=25. Positive number.
    (-5)(-5)=25. Again, positive number.


    Maybe that is the only idea you were missing.
     
  15. Aug 17, 2015 #14

    jbriggs444

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    When taking the square root of a positive real number, the radical sign means the principle square root -- the square root that is positive.

    When taking the square root of a negative real number or of a complex number with a non-zero imaginary part there are two square roots. But neither is a positive real number. There is no way to write down a formula for ##i## using real number arguments that could not equally well be considered to yield ##-i##. The labels ##i## and ##-i## are arbitrary in this sense and could be swapped without changing any mathematics.

    I think that this is what HOI was trying to get at.
     
  16. Aug 17, 2015 #15

    symbolipoint

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    Excellent point, but a little extra work to not be confusing. What you mean is that we have this:
    (-i)(-i)=(-1)(i)(-1)(i)
    (-1)(-1)(i)(i)
    (+1)*i^2
    i^2
    -1
     
  17. Aug 18, 2015 #16

    HallsofIvy

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    thelema418 wrote "I don't understand what you mean when you say this notation is ambiguous. The radical sign means principal square root."
    How do you define "principal square root" for complex numbers?

    Symbolipoint wrote "Exponentiate both sides to the [itex] \frac{1}{2}[/itex] power.
    Plus or Minus Square Root of negative 1; no more ambiguity."
    Exponenting both sides still leaves it ambiguous. Writing "plus or minus" does NOT remove that ambiguity- the complex numbers are NOT an ordered field. We cannot unambiguously designate some as "positive" and others "negative".

    And writing "-i= (-1)i" does not help since the whole question is about distinguishing unambiguously between i and -i.
     
  18. Aug 19, 2015 #17
    Personally, I would use polar coordinates and cis notation. I see it like a piece of paper that has a tear down the non-postive real-axis (arbitrarily). As you move counterclockwise from the non-positive real axis, you eventually loop over to the backside of the paper. Continuing to rotate, you are eventually back where you began. The principal root is what occurs on the front page before you get back to where you started; it is contingent on the aforementioned choices.

    Back to HOI, I do not understand how the argand coordinates resolves notational ambiguity. If ##i = (0, 1)## and ##-i = (0, -1)##, I'm not certain what is different (in the context of ambiguity). Are you really distinguishing between ##i## and ##-i## any differently? Couldn't we just then say ##i = (0, -1)## and ##-i = (0, 1)##? And how is this different than the algebra involved in ##(-i)(-i) = (-1)(i)(-1)(i)##?
     
  19. Aug 19, 2015 #18

    jbriggs444

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    Before you could use polar coordinates, you would have to define which square root of -1 you are going to place above the real axis and which below. How do you know which one gets the label ##i## and which one gets the label ##-i##?
     
  20. Aug 19, 2015 #19
    There is no indicators of where to put 1 and -1 on the real y axis either. The geometry just seems to be part of an assumptive framework to me, much like rotating counterclockwise and locating the cut. You could really do anything.

    This seems like a geometric concern - not a notational ambiguity which is what HOI spoke of. Notationally, ##i \equiv \sqrt{-1}## seems less ambiguous to me than the proposed ##i^2 = -1##. And even saying ##(0, 1) = i## does not define whether the y axis is increasing or decreasing in the downward direction. HOI claimed this was better than the notational ambiguity with ##i \equiv \sqrt{-1}##.
     
  21. Aug 19, 2015 #20

    jbriggs444

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    But -1 and 1 are distinguishable. -1 times -1 is equal to 1. 1 times 1 is not equal to -1. That is not the case for i and -i. The copy of the complex numbers that you get by interchanging the two is isomorphic to the original.
     
  22. Aug 19, 2015 #21
    Yes there are. -1 < 0 < 1 whereas there is no ordering of imaginary numbers.

    Is this because you are clinging to the definition of ## \sqrt x ## as being the positive square root? There is no such thing as a positive imaginary number.

    That's the whole point - the y axis (of the Argand plane) does not increase in any direction. Yet again, there is no ordering of imaginary numbers.

    The important distinction is this: -1 is 1 less than 0; -i is NOT i less than 0.
     
  23. Aug 19, 2015 #22

    pwsnafu

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    The direction of the axis is irrelevant.
    If we define ##i = (0,1)## and ##-i = (0,-1)##, then ##\Re(i) = 1 > 0 > -1 = \Re(-1)##, because the reals are a ordered field and we are using ordered pars of reals in our construction.

    The argument of complex numbers is ##\mathbb{R} / 2\pi##. That is not an ordered field.

    You are describing a two fold covering of the complex numbers, not the complex numbers themselves.That is not what you want to do when you are defining i itself. You can't define i using a two fold covering of the complex numbers without defining the complex numbers to begin with.
     
    Last edited: Aug 19, 2015
  24. Aug 19, 2015 #23

    symbolipoint

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    A geometric description to show the meaning of square root of negative 1 has been derived and a search on YouTube will give some results, at least a few of which will be good. I do not have any specific hyperlink addresses for them.

    One of the motivations for imaginary numbers is the equation x^2+1=0, and to find a solution for what is x.
     
  25. Aug 19, 2015 #24
    I am well aware that there is no ordering of the imaginary numbers.

    For the real numbers, on the y-axis you do not know whether -1 is geometrically above 0 or below 0. In many math classes, students are conditioned to draw the y-axis with numbers increasing upwards. Yet, there is no reason why you could not arrange it the other way. Maybe I am just too fluid with my use of axes that I just do not see this as an issue.

    I am clinging to the concept that ##\sqrt{b}## is a (classical) function. This means that it has a singular (albeit imposed) result. In the case of ##b=-1## we just say we are grabbing one result and labeling it ##i##. The additive inverse of ##\sqrt{b}## is also a solution to ##x^2 = b##. To me it seems more appropriate to investigate properties at that point. Such as finding out that the additive inverse is also the associate -- ordering properties -- etc.

    Maybe it is just for the indistiguishablenesss that I am failing to see ambiguity with this.
     
    Last edited: Aug 19, 2015
  26. Aug 19, 2015 #25
    Yes, I agree. I was asked how to define a principal root of a complex number. I think I'm just going to persist with my belief that this is just an assumptive framework -- as I really have not seen evidence of notational ambiguity (in my opinion). But thank you.
     
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