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Rearrange Euler's identity to isolate i

  1. Apr 4, 2014 #1
    1. The problem statement, all variables and given/known data

    Maybe this is not possible because i does not represent anything quantile and is merely abstract? I'm not sure and maybe you guys can help!


    2. Relevant equations

    [tex] e^{i \pi} + 1 = 0[/tex]

    3. The attempt at a solution

    [tex] e^{i \pi} + 1 = 0[/tex]

    [tex] e^{i \pi} = -1[/tex]

    You cannot take natural log of a negative number so where do I go from here?

    [tex]ln(e^{i \pi})=ln(-1)[/tex]

    [tex]i \pi=ln((-1))[/tex]

    [tex]i=\frac{ln(-1)}{\pi}[/tex]
     
  2. jcsd
  3. Apr 4, 2014 #2

    micromass

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    You would have to take the complex logarithm, which is a subtle little thing.

    https://www.physicsforums.com/showthread.php?t=637214 [Broken]
     
    Last edited by a moderator: May 6, 2017
  4. Apr 4, 2014 #3
    It is true that ln(-1) = ipi. I don't see anything wrong with what you've said.

    You cannot take the ln of a negative in the reals. We are explicitly not limited to the reals.
     
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