1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Algebraic Manipulation of Euler's Identity Leads to a Strange Result
  1. Algebraic manipulation (Replies: 2)