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Euler's Formula Contradiction?

  1. May 9, 2015 #1

    TheDemx27

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    I've been using euler's formula now more than I have in the past, (using it for circuit analysis stuff), and so its been floating around in my head a bit more.

    Say you have [itex]e^{2πi}=1[/itex] and you take the natural log of both sides.

    [itex]\log_e( e^{2πi})=\log_e(1)[/itex]
    [itex]2πi=0[/itex]
    uhhhhh... :confused:
     
  2. jcsd
  3. May 9, 2015 #2
    The problem is that you are using the real natural logarithm, which is the inverse of the real exponential function [itex]e^x[/itex]. You need to use the complex logarithm.
     
  4. May 9, 2015 #3

    TheDemx27

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    Ah, thankyou.
     
  5. May 9, 2015 #4
    The general equation is ##e^{\pm 2ni\pi}=1##
    The function ##e^{ix}## has periods of ##2\pi##, just as the trigonometric functions have periods of ##2\pi##
    [i.e. although ##sin(2\pi)=sin(0),\ 2\pi\neq0##]
     
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