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Thread: 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 $\displaystyle e^{iz}$.
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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 $\displaystyle e^x$, place x=z to get the expansion for z, and x=iz for the expansion of $\displaystyle e^{iz}$

    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 $\displaystyle z = \cos{x} + i\sin{x}$.

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

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

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

    $\displaystyle = iz$.


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

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

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

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

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

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

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

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

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

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


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


    So $\displaystyle 1 = Ae^{0i}$

    $\displaystyle 1 = Ae^0$

    $\displaystyle A = 1$.



    So we have $\displaystyle z = \cos{x} + i\,\sin{x} = e^{ix}$.
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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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    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 $\displaystyle e^{ix} = \cos{x} + i\sin{x}$, what happens when $\displaystyle x = 1$?


    2. $\displaystyle (1 + i)^i$.

    Write $\displaystyle 1 + i$ in polar form.

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


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

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

    $\displaystyle = e^{-\frac{\pi}{4}}$
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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!

    $\displaystyle e^{a+b}= e^ae^b$, not $\displaystyle e^{ab}= (e^a)^b= (e^b)^a$.
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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???
    $\displaystyle e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4 !}+\frac{x^5}{5!}+....$

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

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

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

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

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

    $\displaystyle =Cosz+iSinz$
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