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Math Help - e^(iz)

  1. #1
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    e^(iz)

    hello...

    how would u prove e^(iz) is equla to cosz+isinz???
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  2. #2
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    Quote Originally Posted by mpl06c View Post
    hello...

    how would u prove e^(iz) is equla to cosz+isinz???
    Using the "power series" representations of the 3 functions
    and seperating the real and imaginary parts of e^{iz}.
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  3. #3
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    yeah... this is only true when z is real though..
    so do i set z:R
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  4. #4
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    Quote Originally Posted by mpl06c View Post
    yeah... this is only true when z is real though..
    so do i set z:R
    From the expansion for e^x, place x=z to get the expansion for z, and x=iz for the expansion of e^{iz}

    the exponent of e doesn't have to be real.
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  5. #5
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    Quote Originally Posted by mpl06c View Post
    hello...

    how would u prove e^(iz) is equla to cosz+isinz???
    Let z = \cos{x} + i\sin{x}.

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

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

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

     = iz.


    Now since \frac{dz}{dx} = iz

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

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

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

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

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

    |z| = e^{ix + C}

    |z| = e^Ce^{ix}

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

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


    Now, let's get evaluate the constant. You should know that when x = 0 then z = \cos{0} + i\sin{0} = 1.


    So 1 = Ae^{0i}

    1 = Ae^0

    A = 1.



    So we have z = \cos{x} + i\,\sin{x} = e^{ix}.
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  6. #6
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    hmm ok...
    just have to quick question...

    1. how do u find e^i

    and 2..
    say you have(1+i)^i

    i understand it all and it gets broken down to

    exp(ilog(1+i))

    but why cant it also be written as (e^logi)*(e^(1+i))
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  7. #7
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    Quote Originally Posted by mpl06c View Post
    hmm ok...
    just have to quick question...

    1. how do u find e^i

    and 2..
    say you have(1+i)^i

    i understand it all and it gets broken down to

    exp(ilog(1+i))

    but why cant it also be written as (e^logi)*(e^(1+i))
    1. Since e^{ix} = \cos{x} + i\sin{x}, what happens when x = 1?


    2. (1 + i)^i.

    Write 1 + i in polar form.

    1 + i = \cos{\frac{\pi}{4}} + i\sin{\frac{\pi}{4}} = e^{\frac{\pi}{4}i}.


    So (1 + i)^i = (e^{\frac{\pi}{4}i})^i

     = e^{\frac{\pi}{4}i^2}

     = e^{-\frac{\pi}{4}}
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  8. #8
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    Quote Originally Posted by mpl06c View Post
    hmm ok...
    just have to quick question...

    1. how do u find e^i

    and 2..
    say you have(1+i)^i

    i understand it all and it gets broken down to

    exp(ilog(1+i))

    but why cant it also be written as (e^logi)*(e^(1+i))
    Because they are not equal!

    e^{a+b}= e^ae^b, not e^{ab}= (e^a)^b= (e^b)^a.
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  9. #9
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    Quote Originally Posted by mpl06c View Post
    hello...

    how would u prove e^(iz) is equla to cosz+isinz???
    e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4  !}+\frac{x^5}{5!}+....

    Cosx=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+....

    Sinx=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}-....

    e^{iz}=1+iz+\frac{(iz)^2}{2!}+\frac{(iz)^3}{3!}+\f  rac{(iz)^4}{4!}+\frac{(iz)^5}{5!}+\frac{(iz)^6}{6!  }+\frac{(iz)^7}{7!}...

    =1+iz-\frac{z^2}{2!}-\frac{iz^3}{3!}+\frac{z^4}{4!}+\frac{iz^5}{5!}-\frac{z^6}{6!}-\frac{iz^7}{7!}+\frac{z^8}{8!}+\frac{iz^9}{9!}-....

    =\left(1-\frac{z^2}{2!}+\frac{z^4}{4!}-\frac{z^6}{6!}+...\right)+i\left(z-\frac{z^3}{3!}+\frac{z^5}{5!}-....\right)

    =Cosz+iSinz
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