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

  1. Mar 14, 2013 #1

    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


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    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!
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