1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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)^(-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

    User Avatar
    Homework Helper

    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:


    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

    User Avatar
    Homework Helper

    The step
    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
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook