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Complex numbers: show that (a^b)^c has more values than a^(bc)

  1. Dec 30, 2013 #1
    1. The problem statement, all variables and given/known data

    Show that (a^b)^c can have more values than a^(bc)

    Use [(-i)^(2+i)]^(2-i) and (-i)^5 or (i^i)^i and i^-1 to show this.


    2. Relevant equations



    3. The attempt at a solution

    I'm writing out the second one as the first one is long:

    i^i = e^ilni

    lni = i ([itex]\pi[/itex] /2 ± k2[itex]\pi[/itex])

    so then

    i^i = e^(=[itex]\pi[/itex]/2 ±k2[itex]\pi[/itex])

    (i^i)^i = e^i([itex]\pi[/itex]/2 ± k2[itex]\pi[/itex])

    as e^(±ik2[itex]\pi[/itex]) = 1

    you get (i^i)^i = e^i [itex]\pi[/itex]/2

    which is one value only (equal to i^-1 which is -i)

    this is not what the question suggests

    :confused:

    please help!
     
    Last edited: Dec 30, 2013
  2. jcsd
  3. Dec 30, 2013 #2
    I don't see why you calculated ##(i^i)^i##. The question in the OP asks you to calculate something entirely different.
     
  4. Dec 31, 2013 #3

    haruspex

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    It does? How?

    Applestrudle, I can't see a flaw in what you did. i tried the other pair of expressions and also drew a blank.
     
  5. Dec 31, 2013 #4
    I thought is was a strange question. Just to confirm, for complex numbers the same laws for real numbers for exponents holds right? so for complex a,b,c (a^bc) = (a^b)^b ?
     
  6. Dec 31, 2013 #5

    mfb

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    Neglecting the typo, (a^b)^c can have more ambiguity as it can depend more on the choice of the branch for the logarithm.

    $$(-i)^{2+i} = exp\left((2+i)(-i \frac{\pi}{2} + 2k \pi i)\right)$$
    $$\left((-i)^{2+i}\right)^{2-i} = exp\left((2-i)((2+i)(-i \frac{\pi}{2} + 2k \pi i) +{\color{red}{2l \pi i}})\right) $$
    The red part should give additional solutions.

    Same here:
    (i^i)^i = e^i(π/2 ± k2π ± l 2 π i )
     
    Last edited: Dec 31, 2013
  7. Dec 31, 2013 #6

    D H

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    I assume you meant (ab)c = abc. That identity is valid only for positive real a and real b and c. The ultimate point of this exercise is to show that this identity no longer holds for complex a, b, and c.

    Another one to watch out for is the identity (ab)c = acbc. This, too, is no longer valid for complex a, b, and c.
     
  8. Jan 6, 2014 #7
     
  9. Jan 6, 2014 #8

    mfb

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    Sure you can?
     
  10. Jan 7, 2014 #9
    oh yeah because e is real, i'm new to this sorry.
     
  11. Mar 21, 2014 #10
    what if a and b are real but c is complex?

    For (ab)c = acbc if a and b are real but c is complex would it still hold?
     
  12. Mar 21, 2014 #11

    D H

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    No. That identity is valid if a and b are non-negative and c is real. All bets are off otherwise.
     
  13. Mar 21, 2014 #12
    are you sure you don't mean a and b are real and c in non-negative?
     
  14. Mar 21, 2014 #13

    mfb

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    I think "non-negative" implies real, and negatice c are fine if a and b are (real) positive numbers.
     
  15. Mar 21, 2014 #14

    D H

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    No. I meant what I wrote. Try a=b=-1 and c=1/2. It doesn't work. (ab)1/2 is not equal to a1/2b1/2.
     
  16. Jan 13, 2015 #15

    SBP

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    How does the red part come into play $$2l\pi i$$?
    $$((-i)^{(2+i)})^{(2-i)} = e^{((2-i)((2+i)(-i\pi/2 + 2k\pi i) +2l\pi i))}$$
     
  17. Jan 13, 2015 #16

    mfb

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    You need the logarithm of -i and the complex logarithm has many branches that differ by 2l pi i where l is an integer.
     
  18. Jan 13, 2015 #17

    SBP

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    Isn't that what the i2kpi is for? Why can you add the second i2lpi when raise it to a complex number?
     
  19. Jan 13, 2015 #18

    mfb

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    It has the same reason. You have to use the a^b = exp(b ln(a)) conversion twice.
     
  20. Jan 13, 2015 #19

    SBP

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    I see. Thanks!
     
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