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Problem in apparent contradiction in Euler's Identity?

  1. Dec 25, 2014 #1
    I've worked with Euler's Identity for physics applications quite a few times, but ran into a "proof" of a contradiction in it, which I can't seem to find a flaw in (the only time I've ever had to do any proofs was in high school). I've derived Euler's equation in two different ways in past classes, so I know it works, but I'm at a bit of a loss here.

    ## e^{i\theta} = cos{\theta} + isin{\theta} ##

    Set ##\theta = 2\pi ##

    ## e^{2\pi i} = cos{2\pi} + isin{2\pi} ##

    ## e^{2\pi i} = 1 ##

    Take the natural log:

    ## ln{e^{2\pi i}} = ln{1} ##

    ## 2\pi i = 0 ##

    ## i = sqrt{-1} = 0 ##

    ## -1 = 0 ##

    I think the problem was in using the natural log up there, but I'm not positive.
     
  2. jcsd
  3. Dec 25, 2014 #2

    lurflurf

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    in general we cannot conclude arguments are equal from the fact function values are
    $$\require{cancel}\mathrm{f}(x)=\mathrm{f}(y)\cancel\implies x=y\\
    \text{for example another related common error}\\
    (-1)^2=1^2\cancel\implies -1=1
    $$
     
  4. Dec 25, 2014 #3
    The logarithm of a complex number is usually defined such that it gives you ##\phi## where ##-\pi \lt \phi \le \pi##. So using this definition, ##\ln e^{2\pi i}=0## and not ##2\pi i##. You have assumed that taking the logarithm simply gives you the exponent you had in the beginning, but this is untrue. Using complex numbers, there is no such general function as "getting the exponent", like there is not a general function of getting ##x## back from ##x^2##. One could also say that ##(-1)^2=1^2## therefore ##-1=1##, but this is wrong.

    You are basically saying that ##e^{0}=e^{2\pi i}=e^{4\pi i}=e^{6\pi i}## and therefore ##0=2 \pi i= 4 \pi i=6 \pi i## and so on, but this is untrue. Because all the exponents give the same answer, how should the logarithm function know which exponent you want or which one you had at the beginning? It cannot know that, therefore it is defined to give ##-\pi \lt \phi \le \pi## to make it consistent and the ##2 \pi i## value you got from it is false (using the common definition).
     
  5. Dec 25, 2014 #4

    SteamKing

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    The problem is that ln(z) is multi-valued, like many complex functions.

    If z = x + iy = r e , then ln (z) = ln (r) + iθ = ln |z| + i Arg (z)

    http://en.wikipedia.org/wiki/Complex_logarithm

    The 'proof' you listed doesn't seem to account for the fact that the exponential representation of a complex number is not unique.
     
  6. Dec 25, 2014 #5
    Working with that, would it be:

    ## z = e^{2\pi i}, r = 1 ##

    So ## |z| = e^{2\pi i}e^{-2\pi i} = e^{2\pi i - 2\pi i} = e^{0} = 1 ##

    Then ## Ln{(e^{2\pi i})} = ln{|z|} + i Arg(z) = ln{1} + 2\pi i = 2\pi i ##

    Which doesn't seem to be any help, assuming it's actually possible to equate ## Ln{|z|} = ln(1) ## at all. I'm guessing this is why chingel above mentioned a domain restriction.

    I took a math methods class last semester, and somehow managed to forget the section on complex variables altogether. I got a similar formula from my notes there:

    ## Ln{|z|} = ln{|z|} + i({\theta + 2\pi k}) ##, where the case k = 0 is called the principle logarithm. The notes don't say anything about a domain restriction, so I'm guessing this formulation is the "Riemann surface" method. That makes it pretty obvious that a complex logarithm isn't single-valued. Since it's multi-valued, and ## ln 1 ## is single valued, can we simply not equate the natural log of the complex number with ln(1) at all? This doesn't seem quite right either, as I can imagine z = 1 + i(0), and do:

    ## Ln(1) = ln(1) + i(0 + 2\pi k) ##
     
  7. Dec 25, 2014 #6

    SteamKing

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    Yes, just as e2πi = 1, so too can one say that e2kπi = 1, for k = 1, 2, 3, etc.

    If i2 ≡ -1, we can't conclude that because e2πi = 1 and log (e2πi) = 2πi = log (1) = 0, which would lead to the contradiction that i2 = 02 = 0, which it clearly is not, rather than i2 = -1.

    Anytime i pops up, you can't handle the arithmetic the same as you do with real numbers.
     
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