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Algebraic Manipulation of Euler's Identity Leads to a Strange Result

  1. Apr 9, 2008 #1
    I was playing around with Euler's identity the other day. I came across something that seems contradictory to everything else I know, but I can't really explain it.

    I started with

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

    I rewrote this as

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

    Multiplying by a constant, we have

    [tex]kln[-1] = ki\pi[/tex]

    and using log properties I arrived at

    [tex]ln[-1^{k}] = ki\pi[/tex]

    Now if I set k equal to any even number, I have

    [tex]ln[1] = 0 = ki\pi[/tex]

    This seems to imply that [tex]i\pi[/tex] is 0, however it is not. Furthermore, any even value of k gives the same answer. Why is this?
     
  2. jcsd
  3. Apr 9, 2008 #2
    ln(-1) = (2/2)*ln(-1) = (1/2)*2*ln(-1) = (1/2)*ln(-1^2) = (1/2)*ln(1) = (1/2)*0 = 0

    Therefore, ln(-1) = 0.

    In order for properties of logs to work, you must be taking the log of a valid number. You can not take the log of a negative number, therefore, when you apply the properties of logs to something that is undefined, you get ludicrousness.
     
    Last edited: Apr 9, 2008
  4. Apr 9, 2008 #3
    In the complex domain, log is multivalued, which is to say that it's not even a function in the first place.

    You can read more about it here.
     
  5. Apr 9, 2008 #4
    Thanks!
     
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