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Sqrt (-1) * sqrt (-1)

  1. Jan 29, 2012 #1
    sqrt(A)*sqrt(B)=sqrt(A*B) Therefore,

    sqrt(-1)*sqrt(-1)=sqrt(-1*-1)
    =sqrt(1)
    =1

    This doesn't make sense.Did I do something wrong ?
     
  2. jcsd
  3. Jan 29, 2012 #2

    Simon Bridge

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    In general, [tex]\sqrt{a^2}=\pm a[/tex]... so:[tex]\sqrt{a}^2=\sqrt{a}\sqrt{a}=a=\pm\sqrt{a^2}[/tex]... you have used only the positive square root and a can be positive or negative. But what if a can only be positive? Then:
    [tex]\sqrt{-a}^2=\sqrt{-a}\sqrt{-a}=-a=-\sqrt{a^2}[/tex]... because we've excluded one of the possible results from the last term.
     
  4. Jan 29, 2012 #3

    micromass

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    No. The square root only gives out one value: the positive value. You act like there are two solutions, a positive and a negative. There are not.

    The problem is that [itex]\sqrt{-1}[/itex] isn't well-defined. The square root operator is only defined for positive real numbers, not for negative numbers.

    Furthermore, even if we were to define it, then [itex]\sqrt{ab}=\sqrt{a}\sqrt{b}[/itex] wouldn't hold anymore for negative a and b.
     
  5. Jan 29, 2012 #4
    The confusion between the following is common:

    [tex] a = k^{2} [/tex]
    [tex] k = \pm \sqrt{a} [/tex]

    Here a must be positive (or zero), but k can be anything.
    and [tex] \sqrt{s^{2}} = \mid s \mid [/tex]
     
    Last edited: Jan 29, 2012
  6. Jan 29, 2012 #5

    HallsofIvy

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    The problem is not well defined because, as micromass says, [itex]\sqrt{-1}[/itex] is not well-defined. You will sometimes see "i" defined as [/itex]\sqrt{-1}[/itex], but that simply is not a very good definition- because we don't know what [itex]\sqrt{-1}[/itex] means until after we have defined i. You will sometimes see i defined as "the complex number such that [itex]i^2= -1[/itex]" but that also is not a good definition. Just as in the real number such an an equation has two solutions but that "definition" doesn't specify which i is. With the real numbers, we can specify that [itex]\sqrt{a}[/itex] is the positive number, x, such that [itex]x^2= a[/itex] but with the complex numbers, we cannot do that- the complex numbers does not form an "ordered field".

    What we can do is define the complex numbers as the set of ordered pairs, (a, b) with addition defined by (a, b)+ (c, d)= (a+c, b+d) and multiplication defined by (a, b)*(c, d)= (ac- bd, ad+ bc). Then we can associate the real number a with the pair (a, 0) (that's and "isomorphism" from the ordinary real numbers to the subset of the complex numbers of the form (a, 0)) while defining "i" to be the pair (0, 1): i^2= (0, 1)^2= (0,1)*(0,1)= (0*0- 1*1, 1*0+ 0*1)= (-1, 0)=> the real number -1. Now we can write (a, b)= (a, 0)+ b(0,1)= a+ bi.
     
  7. Jan 29, 2012 #6

    Simon Bridge

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    Aren't there? surely √a2 must be one of +a or -a ... eg, if a=-1 then√(-1)2 = -(-1) or did I need set notation to do this properly?

    ... it is sloppy notation. You know how it arises though?

    a=√a2 is true for a ≥ 0 and -a=√a2 is true for a ≤ 0 isn't it?

    ;)

    What are possible values of a, given b and b=a2?
    I would have put: c=√b ... but since |a|=√b, then a=±c [1]

    ... there are two possible values of a which could give result c.

    Thus it seems quite reasonable[2] to write:

    a=±√a2

    hasn't the surd operation destroyed some of the information about a?
    don't we need to reflect that in our notation?

    of course, that's how we find a given a2.
    the surd returns a single value, but a could be the negative of that value too.
    so the notation is trying to say that there are two possible values of a given only a2.

    Presumably there is a better way to express this?

    ---------------------------
    [1] or should that be: a [itex]\in[/itex] {-c,+c} ?
    [2] though I shall concede that it is better to avoid this sort of construction since a on the LHS is a range of values while the a on the RHS is a single value. Besides, the idea is that we don't have a priori knowledge of a.
     
    Last edited: Jan 29, 2012
  8. Jan 29, 2012 #7

    Deveno

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    the trouble lies with the √ symbol.

    in the real numbers, we can enforce a definition that turns √ into a function (i.e., single-valued).

    but this definition breaks down in the complex numbers. why? because even though the equation:

    z2 = a

    has 2 solutions (unless a = 0), there's no way to to say which one is "the positive one" for all non-zero complex numbers. that is, on the complex plane √ is "multi-valued" (not a function).

    as with the complex logarithm, we have to "choose a branch" before computing a square root (and we have 2 to choose from). so, with complex numbers, it IS NOT TRUE that:

    √z√w = √(zw)

    in particular, it is NOT TRUE that:

    √(-1)√(-1) = √((-1)(-1)) = √1 = 1

    (if we choose the branch where √(-1) = i, we get √(-1)√(-1) = i2 = -1, if we choose the branch √(-1) = -i, we get:

    √(-1)√(-1) = (-i)2 = √1 = -1,

    because THAT branch takes the NEGATIVE square root of a positive real).

    as Halls pointed out, we have TWO choices for √(-1), and there isn't any way to KNOW "which one is the "actual" i. complex conjugation is a field automorphism that fixes the reals, so EITHER choice leads to an algebraic entity that "behaves the same".

    to make an analogy, in a parallel universe, there might exist an earth that uses the yx-plane, instead of the xy-plane. to our way of looking at things, they do stuff "upside-down" and "backwards", and from their perspective, so do we. who's right?

    to make a long story short: the notation √z is ILL-DEFINED, for complex numbers, without further elaboration about what convention you are using. there are similar problems with the so-called "Arg" function, using polar coordinates. you HAVE to specify your intended "angle range".

    yes, it's perhaps a bother that the complex exponential is periodic in i. we expect it to be monotonic, like the real exponential, but 'taint so. so logarithms, and roots (the two are intimately bound up, it turns out) don't behave the same way as we expect from our experience with reals. that means certain rules, which we accept as fact in the real numbers, suddenly turn out not to be true in the complex numbers. this is one of those times.
     
  9. Jan 29, 2012 #8

    Char. Limit

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    Sigh. One of these again, huh? Maybe my signature needs another line.

    The square root operator, √, is defined as the positive square root, i.e. k = √(a) is the positive value such that k^2 = a. The POSITIVE value. Okay, that's one issue solved. One done and I'm on to the next one.

    [itex]\sqrt{a} \sqrt{b} = \sqrt{a b}[/itex] only holds if BOTH a and b are POSITIVE. This is not true in your example. Therefore, you've proven nothing except that there's a reason that it only holds when a and b are positive.
     
  10. Jan 29, 2012 #9

    Deveno

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    with all due respect, √z can be defined on the complex numbers (provided you jump through a few hoops, first....can i hold the hoops? pretty please?), even though no such analogous definition of "positive" exists. but you need to be more specific about what you MEAN by √z (or, to put it another way, "which" angle you're taking "half of", when applying that de moivre business).
     
  11. Jan 29, 2012 #10

    Char. Limit

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    All of this is true, but it's hardly relevant to the OP's problem - and I doubt the OP is thinking on that level anyway.
     
  12. Jan 30, 2012 #11

    chiro

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    Hey rischch and welcome to the forums.

    When you get this kind of thing, it helps to use exponential forms of numbers rather than relying on things that can cause this kind of grief.

    The exponential form of a unit length complex number if given by e^(ix) = cos(x) + i sin(x). You can scale the number by multiplying by a positive constant (i.e. r e^(ix)). Then if you want to do things like multiply SQRT(-1) x SQRT(-1) = i * i, if you do this with exponentiation you get e^(i x pi/2) x e^(i x pi/2) = e^(i x pi) = cos(pi) + i sin(pi) = -1. (since sin(pi) = 0 and i x 0 = 0)

    This method doesn't fail and its useful to look at because it will help you understand how numbers work when you do all kinds of things geometrically.
     
  13. Jan 30, 2012 #12

    Deveno

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    fair enough. suffice to say, the OP was doomed at line one where he (she?) writes:

    sqrt(A)*sqrt(B)=sqrt(A*B)

    because this isn't always true.

    explaining WHY it isn't true, perhaps is a bit more of a can of worms then the OP expected, although it certainly makes the historical suspicion of complex numbers easier to understand (they're not just "the real numbers, but even better").

    again, i want to point out that even using "polar form" (which is what you are doing here), still requires a little caution. for example, although it seems obvious that "half of a 0 angle is 0", one needs to remember that an angle of pi is also a possibility (because of the periodicity in i of the complex exponential). you'll get different square roots of cos(5pi/4) + i sin(5pi/4) and cos(-3pi/4) + i sin(-3pi/4) using de moivre's formula, even though both represent the same complex number (the first will be half the angle back clockwise towards the positive x-axis, and the second will be half the angle back counter-clockwise towards the positive x-axis).
     
    Last edited: Jan 30, 2012
  14. Jan 30, 2012 #13

    morphism

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    I think it's extremely relevant to the OP's problem. Your definition of √a as "the positive k such that k^2=a" only makes sense for positive a. Part of the OP's problem was that they naively applied √ to negative numbers. Deveno's very helpful post explained that while you can indeed think about square roots of negative (and, more generally, complex) numbers, you have to be much more cautious, or otherwise you run into problems as the OP has.
     
    Last edited: Jan 30, 2012
  15. Jan 30, 2012 #14

    Char. Limit

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    True.
     
  16. Jan 30, 2012 #15

    chiro

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    I should have said this explicitly, but I am talking about doing standard arithmetic with complex numbers which includes addition, subtraction, multiplication and division (if you can do this).

    The exponential form will never fail for these operations no matter what numbers you are using (unless you try to divide by a zero modulus complex number).

    If you are trying to do exponentiation of some kind (or an operation associated with it) like trying to find roots for example, then that will require the appropriate machinery.

    Again I apologize for not making the suggestion explicit, but I still stand by the suggestion that when we are talking about complex numbers in the context of arithmetic, the exponential method will not fail.
     
  17. Jan 30, 2012 #16
    In particular, if you believe that, for [itex]z \in \mathbb{C}[/itex],[tex]\begin{align*}z^a \cdot z^b &= z^{a+b} \qquad \mbox{and}\\ \left(z^a\right)^b &= z^{ab}\end{align*}[/tex]then[tex]z^a \cdot z^a = z^{a+a} = z^{2a} = \left(z^2\right)^a[/tex]No?
     
  18. Jan 30, 2012 #17

    Char. Limit

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    If you believe that. I don't believe that the things you believe hold for any real a and b and complex z. In particular, I think we've already shown it failing for a = b = 1/2.
     
  19. Jan 30, 2012 #18

    Deveno

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    in a word: no.

    writing complex numbers as re just moves the problem from the function at hand to the "arg" function, and you STILL have to choose a branch. why? because:

    e = cos θ + i sin θ

    and the trigonometric functions are periodic (this means they are NOT 1-1), so "θ" is not uniquely defined. no matter what "range" you pick, somewhere on a circle, arg is going to have a discontinuity.

    and it's easy to pick examples where z starts out in your specified range, but z2 (using the "standard exponential formula") is not in your range, and you have to "re-calculate" the argument by adding or subtracting 2π.

    here is an explicit example: suppose one (as is typical in engineering) limits the angle range to (-π,π].

    then if [itex]z = \cos(2\pi/3) + i\sin(2\pi/3)[/itex], for z2 you DON"T just "double the angle", you double the angle and subtract 2π.

    or, if you limit the angle range to [0,2π), then when you square [itex]\cos(3\pi/2) + i\sin(3\pi/2)[/itex] you have a similar situation.

    we have to be aware of these subtleties when working with complex numbers, the polar representation is not unique (there's a mod 2π in there somewhere). this rears its ugly head with functions that "undo":

    roots (which "undo powers"), related to:
    logarithms (which "undo exponentiaion"), related to:
    exponentiation (complex exponentiation, like zw, invokes the log function).

    it's not correct, and terribly unfair to someone just starting out studying complex numbers, to say something like "you can't go wrong with the exponential method". yes, you can.
     
  20. Jan 30, 2012 #19

    micromass

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    Nice answers here.

    I would like to get some feedback from the OP on whether we are really answering his question and on whether there are still things unclear in the answers.
     
  21. Jan 30, 2012 #20

    Simon Bridge

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    Agreed, it's time to hear from OP!
    So far it has been a "make naive statement and sit back and watch the egg-heads argue" type of thing. lulz.
     
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