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Math Help - exp(iat+b) = cos(at+b) + isin(at+b) ???

  1. #1
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    exp(iat+b) = cos(at+b) + isin(at+b) ???

    exp(iat+b) = cos(at+b) + isin(at+b)

    pls i need 2 know
    Last edited by mr fantastic; November 26th 2009 at 03:23 AM. Reason: Copied post title into main body of post
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  2. #2
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    Do you mean

    Prove e^{i(at+b)}= \cos{(at+b)}+i\sin(at+b)

    ??
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  3. #3
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    OH its i(at+b) kk thx man
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  4. #4
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    To make the question easier make x=at+b

    And let z = \cos{x} + i\sin{x}.

    It is also important to note that when x = 0, z = 1.

    Then

    \frac{dz}{dx} = -\sin{x} + i\cos{x}

     = i^2\sin{x} + i\cos{x}

     = i(\cos{x} + i\sin{x})

     = iz.


    So \frac{dz}{dx} = iz.


    Rearranging and integrating gives

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

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

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

    \ln{|z|} + C_2 = ix + C_1

    \ln{|z|} = ix + C, where C = C_1 - C_2


    |z| = e^{ix + C}

    |z| = e^Ce^{ix}

    z = \pm e^C e^{ix}

    z = Ae^{ix}.


    Now to find the constant, recall that when x = 0, z = 1.

    So

    1 = Ae^{0i}

    A = 1.


    Therefore

    z = e^{ix}.


    Thus this proves Euler's Formula:

    z = \cos{x} + i\sin{x} = e^{ix} \Rightarrow \cos{at+b} + i\sin{at+b} = e^{i(at+b)}

    *Thanks to Prove it for the code
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