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Question regarding imaginary numbers and euler's formula

  1. Aug 26, 2012 #1
    So, I was thinking about Euler's formula, and I noticed something interesting. Based on the fact that [tex]e^\frac{i\pi}{2} = 1 [/tex], it seems as though [tex]\frac{i\pi}{2} = 0[/tex]. However, this doesn't make any sense. Not only can I not see how this expression could possibly equal 0, but that would imply that [tex]i\pi = 0[/tex] which would in turn imply that [tex]e^{i\pi} = 1[/tex] when it, of course, is equal to -1. At the moment, I have only a very basic understanding of complex/imaginary numbers and their properties, but it seems to me that the implication here is that [tex]\ln(1)[/tex] is not uniquely equal to zero. Is there something I'm missing that shows that this is not the case? If I am correct in this conclusion, is this because of some property of imaginary numbers that I don't know about yet?
     
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  3. Aug 26, 2012 #2

    jedishrfu

    Staff: Mentor

  4. Aug 26, 2012 #3
    Oh dear, how embarassing. What a silly mistake.

    However, now that I think about it, while that may have been a silly arithmetic mistake, it is true that [tex]e^{2i\pi} = 1[/tex] which leads to essentially the same question I had originally.
     
  5. Aug 26, 2012 #4
    Oh dear, how embarassing. What a silly mistake.

    However, now that I think about it, while that may have been a silly arithmetic mistake, it is true that [tex]e^{2i\pi} = 1[/tex] which leads to essentially the same question I had originally.
     
  6. Aug 26, 2012 #5
    Euler's formula says that [itex]e^{ix} = \cos x + i \sin x[/itex].

    Let [itex]x = 2 \pi n[/itex], where [itex]n[/itex] is any integer. Then [itex]e^{ix} = \cos x + i \sin x = 1 + 0i = 1[/itex].

    Therefore, [itex]e^{ix} = 1[/itex] has an infinite number of solutions, all of the form [itex]x = 2 \pi n[/itex]. The two you brought up, [itex]x=0[/itex] and [itex]x = 2 \pi[/itex] are just the ones where [itex]n = 0[/itex] and [itex]1[/itex] respectively.
     
  7. Aug 26, 2012 #6

    D H

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    Staff Emeritus
    Science Advisor

    Correct. The complex logarithm is a multi-valued function. (Aside: "multi-valued function" is a bit of a misnomer, as is "red herring". Red herrings typically are neither red nor are they fishes. Multi-valued functions such as the inverse sine and the complex logarithm are not "functions".)

    When you learn a bit more about complex analysis you'll run into the concept of branch points, branch cuts, and principal values. To make the complex logarithm a true function, one has to (somewhat arbitrarily) choose exactly one of the infinite number of values v that satisfy [itex]\exp(v) = z[/itex] as the "principal value" [itex]v=\textrm{Log}\,z[/itex]. The typical choice is that [itex]-\pi < \arg v \le \pi[/itex]. This choice makes [itex]\textrm{Log}\,1 = 0[/itex], which is the obvious choice. But what about [itex]\textrm{Log}\,(-1)[/itex]? Given Euler's identity [itex]\exp \pi i + 1 = 0[/itex], one "obvious" choice is [itex]\textrm{Log}\,(-1) = \pi i[/itex].

    There's a minor problem here. For small real ε > 0, this choice makes [itex]\textrm{Log}(-1+\epsilon i) \approx \pi i[/itex] but it makes [itex]\textrm{Log}(-1-\epsilon i) \approx -\pi i[/itex]. There's a discontinuity in the neighborhood of z=-1 (or in the neighborhood any other negative real number). That's what you get with branch cuts.
     
    Last edited: Aug 26, 2012
  8. Aug 26, 2012 #7
    Very interesting - thanks for the info!
     
  9. Aug 26, 2012 #8
    Euler's formula is

    [itex]e^{ix} = cos(x) + i\ sin (x)[/itex]

    With that formulation, it's easy to see that [itex]e^{ix}[/itex] must be periodic with period 2pi. Once you get used to thinking about the complex exponential that way, it gets much easier to visualize.
     
  10. Aug 26, 2012 #9
    Ah, I hadn't thought about the periodicity before. Thanks
     
  11. Aug 31, 2012 #10
    The cool proof that I saw goes like this:
    Take the Maclaurin series for eu. That is:
    [tex]e^{u}=1+u+\frac{u^{2}}{2!}+\frac{u^{3}}{3!}+\frac{u^{4}}{4!}+\frac{u^{5}}{5!}+\frac{u^{6}}{6!}+\frac{u^{7}}{7!}+\ldots[/tex]
    Now, say u=a+bi. Then eu=ea+bi=eaebi. We are interested in the ebi part, so let's see what that comes out to:
    [tex]e^{bi}=1+(bi)+\frac{(bi)^{2}}{2!}+\frac{(bi)^{3}}{3!}+\frac{(bi)^{4}}{4!}+\frac{(bi)^{5}}{5!}+\frac{(bi)^{6}}{6!}+\frac{(bi)^{7}}{7!}+\ldots[/tex]
    Now remember that i2 is -1, so we substitute accordingly:
    [tex]e^{bi}=1+bi-\frac{b^{2}}{2!}-\frac{ib^{3}}{3!}+\frac{b^{4}}{4!}+\frac{ib^{5}}{5!}-\frac{b^{6}}{6!}-\frac{ib^{7}}{7!}+\ldots[/tex]
    Hmm, what happens if we split this into two components, so that ebi=f(b)+i*g(b). We get:
    [tex]f(b)=1-\frac{b^{2}}{2!}+\frac{b^{4}}{4!}-\frac{b^{6}}{6!}+\ldots[/tex]
    [tex]g(b)=b-\frac{b^{3}}{3!}+\frac{b^{5}}{5!}-\frac{b^{7}}{7!}+\ldots[/tex]
    Wait! Now, f(b) is the Mclaurin series for cos(b) and g(b) is the Mclaurin series for sin(b). We have, then, that ebi=f(b)+i*g(b)=cos(b)+i*sin(b). Now, plug in b=[itex]\pi[/itex] and we get e[itex]\pi[/itex]i=cos([itex]\pi[/itex])+i*sin([itex]\pi[/itex])=-1.

    I find the computation pretty cool. It is a testament to the power of infinite series :)
     
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