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Fractional Exponent of Negative Numbers

  1. Aug 24, 2012 #1
    I have been thinking on this topic for a while.

    As I have seen on various sites:
    1) a^(b/c) means "c"th root of "a^b".
    2) Also an even root of a negative number does not exist in real numbers.

    Then I want to investigate this formula:


    Mustn't it be equal to b=(-4)^(2/4)

    However if we calculate the power first;

    a=square root of [(-4)^1]= Not a real number
    b=4th root of[(-4)^2]= 2

    What do I miss? I have some confussion about decimal exponents but I hope the answer of this question will solve it.
  2. jcsd
  3. Aug 24, 2012 #2
    That specific law of exponents does not work in negative numbers.
    A way to understand this is that your method entails [itex]\sqrt{-a}=(-a)^{2/4}=a^{1/4}[/itex], but clearly [itex]a^{1/4}\cdot a^{1/4}=\sqrt{a}\neq \sqrt{-a}[/itex].
    Your mistake is that you are applying the law [itex](ab)^{c}=a^c\cdot b^c[/itex], which isn't valid for negative a or negative b.
    Last edited: Aug 24, 2012
  4. Aug 24, 2012 #3
    Thanks for your answer.

    But then, how to handle fractional exponent problems:

    For example how to handle this one;

  5. Aug 24, 2012 #4
    The number -2 can be represented in polar form as [itex]2e^{i\pi}[/itex], and hence the problem of exponentiation reduces to evaluating [itex]2^{4/5}e^{4i\pi/5}=2^{4/5}(\cos(4\pi/5)+i\sin(4\pi/5))[/itex].

    Euler's identity is often employed to give the definition of complex exponentiation.
  6. Aug 24, 2012 #5

    I really like your posts, they are succinct (reducing the chance of confusion) and very accurate.

  7. Aug 24, 2012 #6
    Thanks for your answer.

    So, If I want to determine if an negative base has a real root or not, then should I approach it by expressing it as an complex number?

    Couldn't I say if, for example, -2^1.0954 has a real root or not easily?
  8. Aug 24, 2012 #7
    Well sure you could. But you are going to be able to say that based on Millennial's answer.

    Based on his formula, n^(x) where n < 0 will have a real root iff sin(x*pi) = 0. Otherwise it will be a complex number.
  9. Aug 24, 2012 #8


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    Before answering that, let me just add some more info to the previous question regarding (-2)^(4/5).

    Using the complex definitions of exp and log we can write it as,
    [tex]e^{0.8 \log(-2)} = e^{0.8 (\log(2) + i (1 + 2k) \pi))}[/tex]

    So we see that if take this expression over the complex numbers, then it's multi-valued. Basically we get a bunch of different answers, for k = 0, 1, 2, etc. It's easy to see however, that in this case they repeat after k=4, so there are 5 unique values. We also note that for k=2 we get a purely real solution (since 0.8 * i 5 pi = i 4 pi).

    In general, say we have (-x)^(p/q) where "x" is a positive real and fraction p/q is reduced to it's simplest form. Then we can write it as

    [tex]e^{\frac{p}{q} \log(-x)} = e^{\frac{p}{q} (\log(x) + i (1 + 2k) \pi))}[/tex]

    Here we can see that if "q" is odd then there must be some value "k" for which [itex]1+ 2k = q[/itex], and the corresponding value will be real. Conversely we can show that for "q" even there will be no real valued solutions.
  10. Aug 24, 2012 #9
    Complex exponentiation is not multi-valued, complex logarithm is. It is just that complex exponentiation gives the same answer for different numbers, to be precise, [itex]p^q=e^{p\log(q)}[/itex] is equal to [itex]\left(p+\frac{2\pi in}{\log(q)}\right)^{q}=e^{p\log(q)+2\pi in}[/itex] for [itex]n\in \mathbb{Z}[/itex] through the periodicity of trigonometric functions. However, since complex exponention is not a bijection, complex logarithm is multivalued, if [itex]\log(z)=w[/itex] holds, then for [itex]n\in \mathbb{Z}[/itex], [itex]\log(z)=w+2\pi in[/itex] also holds.

    Just wanted to fix that error.

    Also, if you don't know it, complex logarithm on the principal branch is defined by [itex]\log(z)=\log|z|+i\text{arg}(z)[/itex]. It is easily derivable from the polar form of a complex number. Of course, through my last argument above, a general form that covers not just the principal branch but all branches is given by [itex]\log(z)=\log|z|+i\text{arg}(z)+2\pi in[/itex] for [itex]n\in \mathbb{Z}[/itex].

    I'd like to thank you for the complement :)
    Last edited: Aug 24, 2012
  11. Aug 24, 2012 #10
    Firstly thanks all for your wise answers.

    It is clearer to me now.

    I was thinking about "the simplest form". I have got the logic now. But as I am not good at complex numbers it will take a while to understand why it is required to be simpliest form. I think I can find it after some study.

    Thank you all again for your great help.
  12. Aug 24, 2012 #11
    In complex numbers, the rule:
    (z_1 \, z_2)^{\alpha} \stackrel{?}{=} z^\alpha_1 \, z^\alpha_2
    does not hold in general when α is not an integer. The reason behind this is because non-integer powers in complex numbers are defined through the principal branch of the complex logarithm (Log):
    z^\alpha \equiv \exp \left( \alpha \, \mathrm{Log} (z) \right)
    and, for this branch, the product rule:
    \mathrm{Log}(z_1 \, z_2) \stackrel{?}{=} \mathrm{Log}(z_1) + \mathrm{Log}(z_2)
    does not hold in general. This is because:
    \mathrm{Log} (z) \equiv \ln \vert z \vert + i \, \mathrm{Arg} (z)\,, \ -\pi < \mathrm{Arg} (z) \le \pi
    and the principal arguments (Arg) do not add:
    \mathrm{Arg}(z_1 \, z_2) \neq \mathrm{Arg}(z_1) + \mathrm{Arg}(z_2)\,,
    since they are restricted to the above stated interval.

    For example:
    \vert -4 \vert = 4\,, \ \mathrm{Arg}(-4) = \pi
    (-4)^{\frac{1}{2}} = \exp \left(\frac{1}{2} \, ( \ln 4 + i \, \pi ) \right) = e^{\ln 2} \, e^{i \, \frac{\pi}{2}} = 2 \, i
    On the other hand, [itex](-4)^2 = 16[/itex], and:
    \vert 16 \vert = 16\,, \ \mathrm{Arg}(16) = 0
    (16)^{\frac{1}{4}} = \exp \left(\frac{1}{4} \, ( \ln 16 ) \right)= e^{\ln 2} = 2
    Last edited: Aug 24, 2012
  13. Aug 24, 2012 #12


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    Looking at Millenial's and Dickfore's reply I realize I was wrong to say is was multi-valued, since the complex power is defined using only the principle branch of log. That is, using only k=0 in the equation I wrote.

    Think of this as being a bit like the distinction between the (multiple) solutions to the equation [itex]x^2 = 4[/itex] and and the (single) value of [itex]\sqrt{4}[/itex]. To keep sqrt as a single valued function it is usually defined as only the positive value and we use an explicit plus or minus ([itex]\pm[/itex]) to write all the solutions to the given equation.

    In the same manner we can see a close relation to the value of [itex](-2)^\frac{p}{q}[/itex] and the solutions to the equation [itex]x^q - (-2)^p = 0[/itex]. The equation does have "q" solutions, but [itex](-2)^\frac{p}{q}[/itex] is defined as just the one of these values corresponding to the logarithm principle branch.

    No problems, thanks for that. I hope that the above also helps to clarify the situation to the OP. :smile:
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