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Powers of Complex Numbers

  1. Oct 23, 2009 #1
    I have two ways of evaluating [tex](e^{i 2 \pi}) ^{1/2}[/tex], and they give me different answers. Which one is correct, and why is the other wrong?

    Method 1: [tex](e^{i 2 \pi}) ^{1/2} = e^{i \pi} = -1[/tex]

    Method 2: [tex](e^{i 2 \pi}) ^{1/2} = 1^{1/2} = 1[/tex]
  2. jcsd
  3. Oct 23, 2009 #2


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    Both workings are wrongish, but both answers are right. The problem is that exponentiation is not a function; it's what is sometimes called a "multi-valued function": to each input value, there can be more than one output values.

    If you want to turn it into a function, you have to select a principal value. It's common to define it in terms of Exp and Log, a standard choice of principal value for the exp and log functions.

    Otherwise, you have to learn how to deal with expressions being multi-valued.
  4. Oct 23, 2009 #3
    [tex](a^b)^c[/tex] does not equal [tex]a^{bc}[/tex]. In the case where b and c are integers, they are, but not in general.

    Forget the exponentials and consider

    [tex]((-1)^2)^\frac{1}{2} = 1^\frac{1}{2} = 1[/tex] versus [tex]((-1)^2)^\frac{1}{2} = (-1)^1 = -1[/tex]

    Non-integer exponentiation is messy stuff. Multi-valued expressions can't be properly handled algebraically. Alternatively, using the principal value might not be what the problem is asking for. Wording on the problem is important and you need to keep in mind what each expression "means" at a higher level to avoid silly errors in logic.
  5. Oct 23, 2009 #4
    I'm not sure how to determine the "meaning" of a problem I'm working on. It comes from calculus of residues, where I'm trying to evaluate the integral:

    [tex]\int_0^\infty \frac{x^{\mu-1}}{x + 1} dx .[/tex]

    So, I'm using the complex integral

    [tex]\oint \frac{z^{\mu-1}}{z + 1} dz ,[/tex]

    where the cut line goes from the origin to the right, and the closed contour looks like Pacman eating to the right.

    Clearly, I need to determine the poles of the function that we're integrating. There is only one (simple) pole at [tex]z = -1[/tex]. But, if I use that, I get the wrong answer. But, using [tex]e^{i \pi}[/tex], I get the correct answer (the exponents are turned into a sine function).

    Any ideas how this context reveals whether [tex]-1[/tex] or [tex]e^{i \pi}[/tex] should be used?
  6. Oct 23, 2009 #5
    People say often exponentation cannot be done, but the a good answer would rather be how it can be done and how one should extend the rules of exponentiation in order to make the theory consistent again.

    I don't know the mathematically correct approach. My suggestion is to keep in mind the "hidden one" for exponentiations.
    [tex]a^b=(a\cdot 1)^b=a^b\cdot 1^b=a^b e^{2\pi\mathrm{i}kb}[/tex]
    Therefore for non-integer exponents you always get a multi-value-factor [itex]\exp(2\pi \mathrm{i} bk)[/itex]
    So in both cases your answer will be the set {+1,-1}.

    Not sure exactly how to deal with the contour integral problem :)
    So that I can follow the answers later: can you explain the contour again, i,e. (0,0)->(0,\infty)->???
    Is the integral vanishing on all parts of the contour but the real axis?
    Last edited: Oct 23, 2009
  7. Oct 24, 2009 #6
    I think what you mean to say is that [tex](a^b)^c[/tex] only equals [tex]a^{bc}[/tex] for all a>0, but not necessarily when a<0

    For instance [tex](3^{2.7})^{1.4}[/tex] does equal [tex]3^{2.7 \times 1.4} = 3^{3.78}[/tex]

    Also, when you stated that [tex]((-1)^2)^\frac{1}{2} = (1)^\frac{1}{2} = 1[/tex] you were forgetting that [tex](1)^\frac{1}{2} = \sqrt{1} = \pm1[/tex]
    Last edited: Oct 24, 2009
  8. Oct 24, 2009 #7
    Upon further research and review, I see that [tex](a^b)^c = a^{bc}[/tex] is, in fact, one of the laws of exponents.
  9. Oct 25, 2009 #8


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    No, this is incorrect.

    The [itex]\pm 1[/itex] answer comes from the multi-valued function as stated in this thread already. It does not come from [tex]\sqrt{1}=\pm 1[/tex]
    The square root function is defined to be the principal root, i.e. the positive root only.
  10. Oct 25, 2009 #9
    Both answers are correct, since either answer, squared, gives 1. The function you have quoted allows the user to plot n complex nth roots of 1 on a circle of radius 1, centred at (0,0), on the complex plane.
    For example consider the 4th root of 1. it's easy to see that 1^4=1, so 1 is its own 4th root. But (-1)^4 also =1, so it must be a 4th root of 1 too. The same goes for i^4 and (3i/4)^4, which completes the list of 4th roots of 1.
    Using your function, you can plot the n nth roots of 1.
    I remember how pretty some of the animated patterns were that the roots made when plotted on a pixillated screen, which caused moire effects

    sorry, my computer crashed before I could correct my typo, it's not 3i/4 I meant, but 1^3.
    Last edited: Oct 25, 2009
  11. Oct 25, 2009 #10
    Apparently, your version of the quadratic formula is different from everyone else's. I'm wondering just why is it that they decided to put that pesky "plus or minus" symbol in there anyway? Oh yeah, it's because of the square root function.

    If you need a refresher, look here: http://en.wikipedia.org/wiki/Square_root

    You have that backwards. It's actually: the principal root of the square root function is the positive root only.
  12. Oct 25, 2009 #11


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    If [itex]\sqrt{\ }[/itex] was multi-valued, you wouldn't need the [itex]\pm[/itex]. :tongue:

    Anyways, y'all are quibbling over irrelevancies. [itex]\sqrt{\ }[/itex] is defined precisely how the context defines it -- some authors will prefer to restrict it to positive real arguments only and be single-valued, some authors will restrict it to be the principal square root, and some authors will use it for the two-valued square root.
  13. Oct 26, 2009 #12
    The problem with defining [itex]\sqrt{1}=\pm 1[/itex] is that you have no way of writing if you mean only one particular square root. So I prefer saying
    Now the square root returns only the positive value. So to invert the square take square roots on both sides
    therefore (since the square root and square don't exactly cancel but give the module function instead)
    The solution to this equation is
  14. Oct 26, 2009 #13


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    My version is just fine.

    But apparently, by your version, the quadratic formula is complete simply by [tex]x=\frac{-b+\sqrt{b^2-4ac}}{2a}[/tex] since the square root function already gives us two values. As Hurkyl has already stated, there wouldn't be any point in showing that "pesky [itex]\pm[/itex]" :tongue2:

    I have yet to see any authors use this approach except in complex numbers. It seems like more common practice to give the principal root only when dealing with real numbers and the opposite goes for complex numbers.
    But then again, changing the rules around a bit (which aren't set in stone on this subject, it's just accepted as the norm) could be beneficial to suit each particular situation.
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