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Math Help - Prove e^z = cos (z) + i sin (z)

  1. #1
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    Prove e^{iz} = cos (z) + i sin (z)

    Euler's formula states that e^{\mathbf{i}\theta}=cos(\theta)+\mathbf{i}sin(\th  eta), \ \ \forall\theta\in\mathbb{R}.

    Prove e^{\mathbf{i}z}=cos(z)+\mathbf{i}sin(z), \ \ \forall z\in\mathbb{C}

    This seems trivial but I am going to ask for verification anyways.

    Let  \ z\in\mathbb{C}, \ z=x+\mathbf{i}y, \ x,y\in\mathbb{R}

    e^{\mathbf{i}z}=e^{\mathbf{i}(x+\mathbf{i}y)}=e^{\  mathbf{i}x-y}=e^{-y}(cos(x)+\mathbf{i}sin(x))=e^{-y}e^{x\mathbf{i}}

    =e^{x\mathbf{i}-y}=e^{\mathbf{i}(x+\mathbf{i}y)}=e^{\mathbf{i}z}

    Is it just this simple?
    Last edited by dwsmith; December 21st 2010 at 03:26 PM. Reason: Thread moved to Analysis subforum by MrF.
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  2. #2
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    Your title and the problem do not match. Which is it?

    \cos(z)=\cos(x)\cosh(y)-i~\sin(x)\sinh(y)

    i~\sin(z)=-\cos(x)\sinh(y)+i~\sin(x)\cosh(y)
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  3. #3
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    I forgot the i in e^{iz} in the title.
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  4. #4
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    Quote Originally Posted by dwsmith View Post
    Euler's formula states that e^{\mathbf{i}\theta}=cos(\theta)+\mathbf{i}sin(\th  eta), \ \ \forall\theta\in\mathbb{R}.

    Prove e^{\mathbf{i}z}=cos(z)+\mathbf{i}sin(z), \ \ \forall z\in\mathbb{C}

    This seems trivial but I am going to ask for verification anyways.

    Let  \ z\in\mathbb{C}, \ z=x+\mathbf{i}y, \ x,y\in\mathbb{R}

    e^{\mathbf{i}z}=e^{\mathbf{i}(x+\mathbf{i}y)}=e^{\  mathbf{i}x-y}=e^{-y}(cos(x)+\mathbf{i}sin(x))=e^{-y}e^{x\mathbf{i}}

    =e^{x\mathbf{i}-y}=e^{\mathbf{i}(x+\mathbf{i}y)}=e^{\mathbf{i}z}

    Is it just this simple?
    Your proof seems circular to me. The usual approach is to use power series.
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  5. #5
    MHF Contributor FernandoRevilla's Avatar
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    \cos z+i\sin z=\dfrac{e^{iz}+e^{-iz}}{2}+i\;\dfrac{e^{iz}-e^{-iz}}{2i}=e^{iz}

    Fernando Revilla
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  6. #6
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    Quote Originally Posted by FernandoRevilla View Post
    \cos z+i\sin z=\dfrac{e^{iz}+e^{-iz}}{2}+i\;\dfrac{e^{iz}-e^{-iz}}{2i}=e^{iz}

    Fernando Revilla
    Circular proof.

    The usual way of proving the formulas you have used is to use power series ....
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  7. #7
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    Quote Originally Posted by mr fantastic View Post
    Your proof seems circular to me. The usual approach is to use power series.
    I am little confused about using the power series but here is what I have thought of so far.

    \displaystyle e^{\mathbf{i}}e^{z}\Rightarrow e^{\mathbf{i}}e^{z}=e^{\mathbf{i}}\sum_{n=0}^{\inf  ty}\frac{z^n}{n!}=1+z+\frac{z^2}{2!}+\frac{z^3}{3!  }+\cdots +

    Now, how can I use this to show the e^{i} sum is cos(z)+ i sin(z)??
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  8. #8
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by mr fantastic View Post
    Circular proof.
    No, it is not.

    (i) You can define e^z=e^x(\cos y+i\sin y),\quad z=x+iy\;(x,y\in \mathbb{R}) where \cos y,\;\sin y are the standard \cos and \sin on \mathbb{R}

    (ii) You can define \sin z=\dfrac{e^{iz}-e^{-iz}}{2i} and \cos z=\dfrac{e^{iz}+e^{-iz}}{2} in strict terms of e^{iz} and e^{-iz} which have been already defined

    The usual way of proving the formulas you have used is to use power series ....
    It is a way, but not the only one.

    Fernando Revilla
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  9. #9
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    Quote Originally Posted by dwsmith View Post
    I am little confused about using the power series but here is what I have thought of so far.

    \displaystyle e^{\mathbf{i}}e^{z}\Rightarrow e^{\mathbf{i}}e^{z}=e^{\mathbf{i}}\sum_{n=0}^{\inf  ty}\frac{z^n}{n!}=1+z+\frac{z^2}{2!}+\frac{z^3}{3!  }+\cdots +

    Now, how can I use this to show the e^{i} sum is cos(z)+ i sin(z)??
    e^{iz} \neq e^{i} e^z
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  10. #10
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    Quote Originally Posted by mr fantastic View Post
    e^{iz} \neq e^{i} e^z
    I don't know what I was thinking. I am going to work on this later.
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  11. #11
    MHF Contributor FernandoRevilla's Avatar
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    You can choose:

    First proof:

    \cos z+i\sin z=\dfrac{e^{iz}+e^{-iz}}{2}+i\;\dfrac{e^{iz}-e^{-iz}}{2i}=e^{iz}

    Second proof:

    \cos z+i\sin z=
    1-\dfrac{z^2}{2!}+\dfrac{z^4}{4!}-\ldots
    +i\;\left(z-\dfrac{z^3}{3!}+\dfrac{z^5}{5!}-\right)=
    1+\dfrac{iz}{1!}+\dfrac{(iz)^2}{2!}+\dfrac{(iz)^3}  {3!}+\dfrac{(iz)^4}{4!}+\dfrac{(iz)^5}{5!}+\ldots=  e^{iz}

    Fernando Revilla
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  12. #12
    MHF Contributor Drexel28's Avatar
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    You can also doing this, if you know the theorems:

    A) By appealing to certain facts about analytic continuation (namely, you can prove that your function is true on a certain line [namely the imaginary one] and go from there)

    B) Use (although for this one I'm less sure that it works) particular theorems about the uniqueness to complex valued initial condition equations.
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  13. #13
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by Drexel28 View Post
    A) By appealing to certain facts about analytic continuation (namely, you can prove that your function is true on a certain line [namely the imaginary one] and go from there)
    Right. From

    e^{ix}=\cos x +i\sin x\;(\forall x\in\mathbb{R})

    and using the Milne-Thompson method, we inmediately obtain:

    e^{iz}=\cos z +i\sin z\;(\forall z\in\mathbb{C})

    Fernando Revilla
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  14. #14
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    I liked this question even though I didn't know how to approach it originally.
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