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Math Help - What is wrong?

  1. #1
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    What is wrong?

    I posed to myself two problems:
    1^{\pi}=?
    and
    (-1)^{\pi}=?.
    For the second one I did (-1)^{\pi}=(e^{\pi i})^{\pi}=\cos (\pi^2)+ i\sin(\pi^2) it seems good.
    But for the second one
    1^{\pi}=(e^{2\pi i})^{\pi}=\cos (2\pi^2)+i\sin (2\pi^2) yet it should be one!?!?

    I am guessing the exponent for complex numbers is defined for e^{\theta i} where 0\leq \theta<2\pi Thus, my manipulation in problem 2 is out of the domain of the definition and it causes problems.
    Am I right?
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by ThePerfectHacker
    I posed to myself two problems:
    1^{\pi}=?
    and
    (-1)^{\pi}=?.
    For the second one I did (-1)^{\pi}=(e^{\pi i})^{\pi}=\cos (\pi^2)+ i\sin(\pi^2) it seems good.
    But for the second one
    1^{\pi}=(e^{2\pi i})^{\pi}=\cos (2\pi^2)+i\sin (2\pi^2) yet it should be one!?!?

    I am guessing the exponent for complex numbers is defined for e^{\theta i} where 0\leq \theta<2\pi Thus, my manipulation in problem 2 is out of the domain of the definition and it causes problems.
    Am I right?
    1=e^{2.n.\pi .i}, and -1=e^{(2.n+1).\pi .i}, for n=0, \pm 1, \pm 2, ..

    Now raise these to the power \pi, and you have multiple values one
    of which will probably be what you are expecting.

    RonL
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  3. #3
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    Quote Originally Posted by CaptainBlack
    1=e^{2.n.\pi .i}, and -1=e^{(2.n+1).\pi .i}, for n=0, \pm 1, \pm 2, ..

    Now raise these to the power \pi, and you have multiple values one
    of which will probably be what you are expecting.

    RonL
    Thus, you are saying there are infinitely many answers?
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by ThePerfectHacker
    Thus, you are saying there are infinitely many answers?
    Yes, if we alow complex values.

    RonL
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  5. #5
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    Quote Originally Posted by CaptainBlack
    Yes, if we alow complex values.

    RonL
    Then explain this to me:
    1^{\pi}=(e^{2\pi i})^{\pi}=\cos{2\pi^2}+i\sin{2\pi^2}
    And,
    1^{\pi}=(e^0)^{\pi}=\cos0+i\sin0
    Thus,
    1=\cos0+i\sin0=\cos{2\pi^2}+i\sin{2\pi^2}
    But that is not true ?!?!
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  6. #6
    TD!
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    You have to be careful with complex numbers, not all the properties you know from the real numbes (about square roots, powers, logarithms, ...) still hold for complex numbers.

    In particular, the rule (z^a)^b = z^{ab} no longer holds.
    The correct version is (z^a)^b = z^{ab}e^{2bk\pi i}.

    So (e^{2\pi i})^{\pi} = e^{2\pi^2 i}e^{2k\pi^2 i} = e^{2\pi^2 i+2k\pi^2 i} = e^{(k+1)2\pi^2 i}

    And this again yields e^0 for k = -1.

    This has to do with the analytical continuation of the natural logarithm as a complex function and the fact that for complex values of z, e^z is a periodic function with period 2\pi i.
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  7. #7
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    Quote Originally Posted by TD!
    You have to be careful with complex numbers, not all the properties you know from the real numbes (about square roots, powers, logarithms, ...) still hold for complex numbers.

    In particular, the rule (z^a)^b = z^{ab} no longer holds.
    The correct version is (z^a)^b = z^{ab}e^{2bk\pi i}.

    So (e^{2\pi i})^{\pi} = e^{2\pi^2 i}e^{2k\pi^2 i} = e^{2\pi^2 i+2k\pi^2 i} = e^{(k+1)2\pi^2 i}

    And this again yields e^0 for k = -1.

    This has to do with the analytical continuation of the natural logarithm as a complex function and the fact that for complex values of z, e^z is a periodic function with period 2\pi i.
    Thank you TD!, well answered.
    I knew that all of these "paradoxes" I was getting was the result of incorrect rules for complex numbers. It is like the square root of the product of two negative numbers IS NOT the product of the square roots. I just never studied complex analysis.
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  8. #8
    TD!
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    Quote Originally Posted by ThePerfectHacker
    Thank you TD!, well answered.
    I knew that all of these "paradoxes" I was getting was the result of incorrect rules for complex numbers. It is like the square root of the product of two negative numbers IS NOT the product of the square roots. I just never studied complex analysis.
    Indeed, that's also one of the consequences. It follows from the fact that when you analytically continue the square root function as a complex function, you'll have to introduce a branch cut and this makes it possible for the 'old rule' we had for the real case to give false results in the complex case.
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  9. #9
    Grand Panjandrum
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    Quote Originally Posted by ThePerfectHacker
    Then explain this to me:
    1^{\pi}=(e^{2\pi i})^{\pi}=\cos{2\pi^2}+i\sin{2\pi^2}
    And,
    1^{\pi}=(e^0)^{\pi}=\cos0+i\sin0
    Thus,
    1=\cos0+i\sin0=\cos{2\pi^2}+i\sin{2\pi^2}
    But that is not true ?!?!
    If you replace the \pi exponent by 1/2 this
    line of argument will give 1=-1

    RonL
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