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Math Help - e ^ i pi = -1? what?

  1. #1
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    e ^ i pi = -1? what?

    im quite baffled by this question i was recently asked
    so why does e^ (i*pi) = -1?

    Any information is greatly appreciated!
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  2. #2
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    Well it is known that \displaystyle e^{i\theta} = \cos \theta +i\sin \theta

    Now make \theta = \pi what do you get?
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    Quote Originally Posted by lochnessmonster View Post
    im quite baffled by this question i was recently asked
    so why does e^ (i*pi) = -1? Any information is greatly appreciated!
    Using infinite series representations for \cos(t)~\&~\sin(t)
    we can show that \displaystyle e^{\mathif{i}t}=\cos(t)+\mathif{i}\sin(t).
    So let t=\pi.
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    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by lochnessmonster View Post
    im quite baffled by this question i was recently asked
    so why does e^ (i*pi) = -1?

    Any information is greatly appreciated!
    Let's suppose that Leonhard Euler has been a poet and not a mathematician, so that we ignore his formula. In that case the answer to Your question may derive from the solution of the differential equation...

    z^{'}= i\ z\ , \ z(0)=1 (1)

    The (1) is the trajectory of a point the velocity of which is its postion rotated by \displaystyle \frac{pi}{2} so that we have an 'uniform circular move'. The solution of (1) is z=e^{i\ t} so that the question is: where will be the point at the time t= \pi?...

    Kind regards

    \chi \sigma
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  5. #5
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    Quote Originally Posted by chisigma View Post
    Let's suppose that Leonhard Euler has been a poet and not a mathematician, so that we ignore his formula. In that case the answer to Your question may derive from the solution of the differential equation...

    z^{'}= i\ z\ , \ z(0)=1 (1)

    The (1) is the trajectory of a point the velocity of which is its postion rotated by \displaystyle \frac{pi}{2} so that we have an 'uniform circular move'. The solution of (1) is z=e^{i\ t} so that the question is: where will be the point at the time t= \pi?...

    Kind regards

    \chi \sigma
    In fact, any complex number can be written as \displaystyle z = r\cos{x} + i\,r\sin{x}

    so it is quite easy to see that \displaystyle \frac{dz}{dx} = -r\sin{x} + i\,r\cos{x}

    \displaystyle = i^2r\sin{x} + i\,r\cos{x}

    \displaystyle = i(r\cos{x} + i\,r\sin{x})

    \displaystyle = iz.


    Since \displaystyle \frac{dz}{dx} = iz

    \displaystyle \frac{1}{z}\,\frac{dz}{dx} = i

    \displaystyle \int{\frac{1}{z}\,\frac{dz}{dx}\,dx} = \int{i\,dx}

    \displaystyle \int{\frac{1}{z}\,dz} = ix + C_1

    \displaystyle \log{|z|} + C_2 = ix + C_1

    \displaystyle \log{|z|} = ix + C where \displaystyle C = C_1 - C_2

    \displaystyle |z| = e^{ix + C}

    \displaystyle |z| = e^Ce^{ix}

    \displaystyle z = Ae^{ix} where \displaystyle A = \pm e^C.


    So \displaystyle z = r\cos{x} + i\,r\sin{x} = Ae^{ix}, and since we know that when \displaystyle x = 0, z = r

    \displaystyle r = A.


    Therefore \displaystyle r\,e^{ix} = r\cos{x} + i\,r\sin{x}.

    Now by letting \displaystyle x = \pi we have

    \displaystyle r\,e^{i\pi} = r\cos{\pi} + i\,r\sin{\pi}

    \displaystyle e^{i\pi} = \cos{\pi} + i\sin{\pi}

    \displaystyle e^{i\pi} = -1 + 0i

    \displaystyle e^{i\pi} = -1

    \displaystyle e^{i\pi} + 1 = 0.


    This is considered to be the most beautiful equation known, because it links the five fundamental constants of mathematics.
    Last edited by Prove It; April 12th 2011 at 02:07 AM.
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    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by Prove It View Post
    ... in fact any complex number can be written as \displaystyle z = \cos{x} + i\sin{x}...
    ... that may be the most 'revolutionary' idea of the 'new millennium'... I suggest however all students to forget that in their school tests...

    Kind regards

    \chi \sigma
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    Quote Originally Posted by chisigma View Post
    ... that may be the most 'revolutionary' idea of the 'new millennium'... I suggest however all students to forget that in their school tests...

    Kind regards

    \chi \sigma
    I've realised that technically speaking that should read \displaystyle z = r\cos{x} + i\,r\sin{x}, but the proof is identical :P
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  8. #8
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    Quote Originally Posted by Prove It View Post
    I've realised that technically speaking that should read \displaystyle z = r\cos{x} + i\,r\sin{x}, but the proof is identical :P
    Technically speaking we should not be leaving error in post just because they are convienient.

    CB
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  9. #9
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    there are many ways of "deriving" this truth. some require you to have some knowledge of complex numbers, or calculus, upon which a sort of proof is based.

    i, however, will not do this. instead, i will try to give you a sense of what e^z MEANS, when z is a complex number.

    one can write a complex number as a two-part sum, one component in the "x" or "real" direction, and another in the "y" or "imaginary" direction. so we can write z as follows:

    z = x + iy. now if we assume that exponents obey some of the rules we have come to expect in the real case, we ought to have:

    e^z = e^(x+iy) = (e^x)(e^iy).

    and so far, so good, and the e^x part behaves as expected, as x gets bigger, it gets bigger. as x becomes very negative, it tends toward 0.

    but the overall effect is not to transform the plane into some inverted parabola-like shape, as you might expect. and this is because of the e^(iy) part.

    if you hold x constant, and travel along the y-direction, e^(iy) traces out a circle in the complex plane. up-and-down, rather than being strictly increasing,

    e^(iy) is periodic, like a trig function. so while on the left-hand side of the y-axis, e^z is rather flat, on the right-hand side it increases rapidly to the left in amplitude,

    and undulates up and down as you travel either up, or down, in a delicate balance of interacting sines and cosines. along the vertical line x = 0,

    e^(iy) is bending the y-axis into a circle of radius 1, starting anew every 2 \pi (which is the period).

    and so it happens, that at y = \pi, we are half-way 'round a circle-cycle, and the value of the combined coordinates is -1 + 0i
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