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

Taking the natural logarithm of e^(2i*pi)

  1. Mar 14, 2013 #1
    Hello,

    I was playing around with DeMoivre's formula

    ei*pi = -1

    and there is something I don't quite understand about taking the natural logarithm of a certain expression. I though that

    e2i*pi = 1

    ln[e2i*pi] = ln (1),

    but this yields to an imposibility

    2i*pi = 0.

    So obviously I am doing something wrong, and when I input ln[e^(2i*pi)] into Wolfram, it gives log[e2i*pi]=0.

    Can anyone explain why Wolfram Alpha translates ln[e^(2i*pi)] to log[e2i*pi]=0, and why that second expression is true?

    Thank you in advance for your time.
     
  2. jcsd
  3. Mar 14, 2013 #2

    WannabeNewton

    User Avatar
    Science Advisor

    https://www.physicsforums.com/showthread.php?t=637214 [Broken] It says micromass but it was really me...>.> ok it was micromass
     
    Last edited by a moderator: May 6, 2017
  4. Mar 14, 2013 #3
    Great, I appreciate, thank you!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Taking the natural logarithm of e^(2i*pi)
  1. Natural logarithms (Replies: 14)

  2. Natural Logarithm Laws (Replies: 7)

Loading...