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

Logaritm identity

  1. Dec 9, 2006 #1
    1. The problem statement, all variables and given/known data

    Show that, for any two nonzero complex numbers z_1 and z_2,
    \text{Log } (z_1 z_2) = \text{Log } z_1 + \text{Log } z_2 + 2 N \pi i \, ,

    where N has one of the values -1, 0, 1.

    2. Relevant equations

    The logarithm on the principal branch is:
    &\text{Log } z = \ln r + i \Theta \, ,\\
    &r > 0 \text{ and } -\pi < \Theta < \pi \, .

    3. The attempt at a solution

    I tried writing z_1 z_2 as exp(log(z_1) + log(z_2)) and taking the log that way, and I ended up getting the result above, but with N being allowed to take on any integer value. Note that
    \log z = \ln |z| + i \arg z

    in general.
    Last edited: Dec 9, 2006
  2. jcsd
  3. Dec 9, 2006 #2


    User Avatar
    Science Advisor
    Homework Helper

    Use the equation you have for log under "attempt at a solution," because that's really the relevant equation. And don't express z1z2 as exp(log(z1) + log(z2)), express it in the form reiq.
  4. Dec 9, 2006 #3


    User Avatar
    Homework Helper

    The point with the N=1,0,-1 business is that when you add two numbers in (-pi,pi), you might not get a number in (-pi,pi), although you will get one in (-2pi,2pi). So, since adding an integer multiple of 2pi to the arg doesn't affect the result, you can either add or subtract 2pi and get back in the necessary (-pi,pi) range.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook