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Thread: e to the i pi misunderstanding

  1. #1
    Mar 2009

    e to the i pi misunderstanding

    Dunno if this should go in this subforum, but I came across it working on a differential equation.

    So e^(xi) = cosx + isinx,
    therefore e^(i*pi) = -1.
    And (-1)^(1/(1+\-2k)) = -1, where k is any integer.

    Now, z^(a*b) = z^(a^(b)) = z^(b^(a)),
    e^(i*pi*(1/3)) = e^((i*pi)^(1/3))

    Since e^(i*pi) = -1,
    e^((i*pi)^(1/3)) = (-1)^(1/3) = -1

    But that isn't correct, I know, because the worked-out solution to this problem from, my calculator, and the answer in the back of my textbook all say to me that e^((i*pi)^(1/3)) = 1/2 + (sqr(3)/2)i.

    My calculator even knows that this -1 obtained from raising e to the i pi isn't the same as a -1 obtained from simply punching it in. I put e^(i*pi) on my calculator, got -1 as an answer, and stored it as A. Then I put in -1, got -1 as an answer, and stored it as B. Then I put in A^(1/3) and B^(1/3), and got 1/2 + (sqr(3)/2)i and -1, respectively.

    (Also, I raised e to pi, then raised that answer to i, and got -1 + 1E-13i, really close to -1, but not actually it. I know that calculators - I have a TI-84 Plus - might do things more arithmetically than a human, just to be simple, and so might obtain slightly different answers than one would by using exact theory, but still...)

    So... e^(i*pi) doesn't REALLY equal -1? What's going on here? What is the reasoning behind why e^(i*pi*(1/3)) = 1/2 + (sqr(3)/2)i (or possibly 1/2 +\- (sqr(3)/2)i)?
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  2. #2
    Moo is offline
    A Cute Angle Moo's Avatar
    Mar 2008
    P(I'm here)=1/3, P(I'm there)=t+1/3

    (-1)^(1/3) = -1
    This equality is false...

    let x be the possible values for $\displaystyle (-1)^{1/3}$

    we have $\displaystyle (-1)^{1/3}=x$
    so $\displaystyle x^3+1=0$

    this factors into $\displaystyle (x+1)(x^2-x+1)=0$
    so $\displaystyle x=-1$ or $\displaystyle x^2-x+1=\left(x-\frac{1-i \sqrt{3}}{2}\right)\left(x-\frac{1+i \sqrt{3}}{2}\right) \Rightarrow x=\frac{1-i \sqrt{3}}{2} \text{ or } x=\frac{1+i \sqrt{3}}{2}$

    more generally, the third roots of the unity, $\displaystyle e^{i \pi/3}~,~ e^{2i \pi/3} ~,~ e^{i \pi}$ satisfy the equation $\displaystyle x^3=-1$

    you could have written :
    $\displaystyle e^{i\pi/3}=\cos \left(\tfrac \pi 3\right)+i \sin \left(\tfrac \pi 3\right)=\frac 12+i \frac{\sqrt{3}}{2}$
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  3. #3
    Mar 2009
    That makes sense.

    That x has three roots if x = (-1)^(1/3) is one of the things I forgot here.

    Thanks, that was a simple and concise explanation.
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