Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Is this correct

  1. Sep 23, 2003 #1

    jcsd

    User Avatar
    Science Advisor
    Gold Member

    Is this correct (I was just a little worried that I've got a power of 1 that's not equal to 1)?:

    1ln(5)/2πi = 5

    proof:

    using x = π in Euelr's identity:

    exi = cos(x) + isin(x)

    eπi = -1

    square both sides:

    e2πi = 1

    putting both sides to the power of ln(5)/2πi


    1ln(5)/2πi = (e2πi)ln(5)/2πi = eln(5) = 5
     
    Last edited: Sep 23, 2003
  2. jcsd
  3. Sep 23, 2003 #2

    Hurkyl

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    In general, the law

    (a^b)^c = a^(bc)

    does not hold in the complex domain.
     
  4. Sep 23, 2003 #3

    jcsd

    User Avatar
    Science Advisor
    Gold Member

    Is there anyway to prove this incorrect, because it seems to me the well-known identity:

    ii = e-π/2

    can be derived from Euler's identity using (a^b)^c = a^(bc), at least suggesting the possibilty that it may hold.

    Of course, using one of the terms in my derivation, you get a patently incorrect answer:

    e2πi = 1 =>

    2πi = ln(1) = 0 =>

    -4π2 = 0
     
  5. Sep 23, 2003 #4

    Integral

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Why must this be incorrect? Remember you are in the complex plane now, not the Reals, much of what you know about the Reals does not apply.
     
  6. Sep 23, 2003 #5

    Tom Mattson

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Because -4p2 does not equal zero, silly!

    That's right. In this case, the thing that does not apply is the inverse relationship between ln(z) and z=reiθ. In the complex plane, it is...

    ln(z)=ln(r)+i(θ+2pk),

    for any integer k. In this case, z=e2pi and so:

    ln(e2pi)=ln(1)+i(2pi+2pk),

    for any integer k. Note that ln(z) is multivalued.

    edit: fixed superscript bracket
     
  7. Sep 23, 2003 #6

    Hurkyl

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    You don't even have to leave the real domain to find counterexamples:

    ((-1)^2)^(1/2) = 1

    but

    (-1)^(2 * 1/2) = -1

    So even in the simpler case, we see that the identity is only valid on certain domains.


    In the complex plane, (the principal value of) the exponential operator is defined to be:

    a^b = exp(b Log a)

    Where Log is the principal value of the logarithm function (the imaginary part is constrained to be in (-π, π]

    So if we inspect the identity:

    (a^b)^c = (exp(b Log a))^c = exp(c Log exp(b Log a))

    while

    a^(bc) = exp(bc Log a)


    In the case of positive real numbers, one uses the fact that ln and e^ are inverse functions to establish the identity... but in the complex domain, exp is not an invertible function and ln is a multivalued function.
     
    Last edited: Sep 23, 2003
  8. Sep 24, 2003 #7

    jcsd

    User Avatar
    Science Advisor
    Gold Member

    Yes, I guessed ln would be a multi-valued function which would invalid the silly result.

    Back to your example: ((-1)^2)^1/2), this is also multi-valued as the answers are -1 and 1 so the relationship at least holds for one of the values.

    I think to prove whether the orginal equality is correct or incorrect I would need to go back to Euler's identities and try to prove it from there.
     
  9. Sep 24, 2003 #8

    jcsd

    User Avatar
    Science Advisor
    Gold Member

    If anyone's intrested, the orginal equality is correct (according to a Maths PhD student anyway)
     
  10. Sep 24, 2003 #9

    Hurkyl

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    My gut tells me you can say something like "The set of possible values of the LHS is a subset of the set of possible values of the RHS" but I haven't worked it out yet.

    (Incidentally, if you use the principal value of the exponentiation function, the original equality is not correct)
     
  11. Sep 24, 2003 #10

    jcsd

    User Avatar
    Science Advisor
    Gold Member

    This is his exact reply for you to see Hurkyl (I sold him a bit short actually, he's a maths lecturer at Princeton):

     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?