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Complex numbers question

  1. May 12, 2012 #1
    1. The problem statement, all variables and given/known data

    Verify that the equation (z1z2)^w=(z1^w)(z2^w) is violated for z1=z2=-1 and w=-i.
    Under what conditions on the complex values z1 and z2 does the equation hold for all
    complex values of w?

    2. Relevant equations



    3. The attempt at a solution

    ((-1)x(-1))^(-i)=1^(-i)=e^ln(1^(-i))=e^(-iln1)=e^0=1

    (-1)^(-i)x(-1)^(-i)=(-1)^(-2i)=((-1)^2)^(-i)=1^(-i) but this is the same as above.

    Thanks in advance for helping. :-)
     
  2. jcsd
  3. May 12, 2012 #2

    I like Serena

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    Hi Lucy Yeats! :smile:

    I'm afraid you can't just use the regular rules for "ln" on complex numbers.
    "ln" is not a well defined function for complex numbers.
    It's a so called multi-valued function.

    You have a similar problem with a^(bc).
    It is not just equal to (a^b)^c.
    It's more complex. ;)


    Try using the length-angle representation of a complex number.
    That is, ##z=r e^{i \phi}##, where ##0 \le \phi \lt 2\pi## and ## r \ge 0##.
    So try writing -1 as ##e^{i \pi}## and 1 as ##e^0##.
     
  4. May 12, 2012 #3
    I know that ln(z1z2)=lnz1+lnz2+2πin, where n is an integer.

    I don't see why I need to use logs in this question. why can't I say:

    (-1x-1)^(-i)=1^(-i)
    (-1)^(-i)x(-1)^(-i)=((-1)^(-i))^2=(-1)^(-2i)=((-1)^2)^(-i)=1^(-i)

    Thanks for helping!
     
  5. May 12, 2012 #4
    Sorry, I see what you mean about the a^bc thing. Ignore the last post and I'll try again.
    :-)
     
  6. May 12, 2012 #5

    I like Serena

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    The step
    (-1)^(-2i)=((-1)^2)^(-i)
    does not hold.

    Try the same step with e^(i pi), which is -1.

    And no, you should not use logs in this question - at least not on complex numbers, but at most only on real numbers.


    EDIT: Too late. ;-)
     
    Last edited: May 12, 2012
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