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Math Help - the i*pi=ln(-1) identity

  1. #1
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    the i*pi=ln(-1) identity

    Famously, we have...

    e^{i \pi (2n-1)}+1=0

    where n is any positive integer. This is of course because \cos{(\pi (2n-1))}=-1 and i\sin{(\pi (2n-1))}=0

    My question is, from this relation that Euler discovered doesn't it also follow that...

    e^{i \pi 2n}-1=0 ?

    Since \cos{(\pi 2n)}=1 and i\sin{(\pi 2n)}=0 ?

    And if this is true, that means that x^{i\pi 2n}=x^{\ln{1}}=x^0=1, right?

    Thanks
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  2. #2
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    Quote Originally Posted by rainer View Post
    Famously, we have...

    e^{i \pi (2n-1)}+1=0

    where n is any positive integer. This is of course because \cos{(\pi (2n-1))}=-1 and i\sin{(\pi (2n-1))}=0

    My question is, from this relation that Euler discovered doesn't it also follow that...

    e^{i \pi 2n}-1=0 ?

    Since \cos{(\pi 2n)}=1 and i\sin{(\pi 2n)}=0 ?

    And if this is true, that means that x^{i\pi 2n}=x^{\ln{1}}=x^0=1, right?

    Thanks
    You are correct that (e^{i\pi})^{2n} - 1 = 0.

    I don't see how you turned i\pi 2n into \ln{1} though... What is x?
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    Quote Originally Posted by Prove It View Post
    You are correct that (e^{i\pi})^{2n} - 1 = 0.

    I don't see how you turned i\pi 2n into \ln{1} though... What is x?
    i\pi 2n=\ln1 follows from euler's equation: e^{i\pi 2n}-1=0 Add 1 to both sides and then take the natural log of both sides and you get i\pi 2n=\ln1

    Right?

    As for x, it is any real number.
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    You cannot go from e^{2ni\pi}= 1 to 2ni\pi= ln(1)= 0 so quickly. In the complex numbers the logarithm is multi-valued. Specifically, if, in polar form, x= re^{i\theta} then it is also true that x= re^{i(\theta+ 2n\pi}. ln(x)= ln(r)+ 2n\pi where n can be any integer. In particular, since 1, in polar form, has r= 1, \theta= 0, ln(1) has many values, one of which is 0. The others are, just as you say, multiples of 2\pi i. It does not follow that, just because as a real function, ln(1)= 0 that 2n\pi i= 0.
    Last edited by HallsofIvy; July 31st 2010 at 01:17 AM.
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    Quote Originally Posted by HallsofIvy View Post
    You cannot go from e^{2ni\pi}= 1 to 2ni\pi= ln(1)= 0 so quickly. In the complex numbers the logarithm is multi-valued. Specifically, if, in polar form, x= re^{i\theta} then it is also true that x= re^{i(\theta+ 2n\pi}. ln(x)= ln(r)+ 2n\pi where n can be any integer. In particular, since 1, in polar form, has r= 1, [itex]\theta= 0[/tex], ln(1) has many values, one of which is 0. The others are, just as you say, multiples of 2\pi i. It does not follow that, just because as a real function, ln(1)= 0 that 2n\pi i= 0.
    Hmmm, so if it's not i\pi 2n=\ln1 what equation would capture this multi-valued character of \ln1 in the complex numbers?

    Also, I do not quite follow your math.

    If x= re^{i(\theta+ 2n\pi)}, then ln(x)= ln(r)+ i(\theta+2n\pi), not ln(x)= ln(r)+ 2n\pi as you wrote. Or is what you wrote somehow the same? What am I missing here?
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  6. #6
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    Sorry, I shifted too quickly to the specific case with \theta= 0.

    In general, if x= re^{i\theta} then it is also true that x= re^{i(\theta+ 2n\pi} so ln(x)= ln(r)+ i(\theta+ 2n\pi)

    In particular, if x= 1, then r= 1, \theta= 0 so ln(x)= 2n\pi for all n.
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    Quote Originally Posted by HallsofIvy View Post
    Sorry, I shifted too quickly to the specific case with \theta= 0.

    In general, if x= re^{i\theta} then it is also true that x= re^{i(\theta+ 2n\pi} so ln(x)= ln(r)+ i(\theta+ 2n\pi)

    In particular, if x= 1, then r= 1, \theta= 0 so ln(x)= 2n\pi for all n.
    Wait, if x=1, and consequently r=1 and theta=0, then what you get is...

    \ln(1)= \ln(1)+ i(0+ 2n\pi)= i\pi 2n not \ln(1)= \pi 2n as HallsofIvy wrote...

    In either case this seems to confirm rather than repudiate the original absurd result I started the thread with via my variation of Euler's identity, viz.:

    e^{i\pi2n}=1

    i\pi2n=\ln(1)=0

    Correct me if I'm wrong, but as far as I can see we're back to square one again. Any other suggestions as to where the fallacy lies in this absurd result?
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    Yes, you are wrong for exactly the reasons stated before: ln(z) is NOT a single valued function. You cannot just assert that ln(1)= 0 in the complex numbers. ln(1)= 2m\pi i for any integer m. Yes, m= 0 gives ln(1)= 0 as one of the values of ln(1) but you cannot, in the complex numbers, start from a= ln(1) and declare that a= 0. If a= ln(1) then a is equal to one of the infinite number of possible values of ln(1).

    Saying that "[itex]2n\pi i= ln(x)[/itex]" only says that [itex]2n\pi i= 2m\pi i[/itex] for some integer m- which is trivially true.
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