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Thread: Prooving DeMoivre's Theorem

  1. #1
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    Prooving DeMoivre's Theorem

    I am having so much trouble prooving this theorem. This is what I have to prove:

    PROVE: If z = r(cos u + i sin u), then z^n = r^n(cos nu + i sin nu) when n = 2

    I am so frustrated at this point. I have to list all my steps and no matter how I try to fix it, it never comes out correctly.

    Please help.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by missyd819 View Post
    I am having so much trouble prooving this theorem. This is what I have to prove:

    PROVE: If z = r(cos u + i sin u), then z^n = r^n(cos nu + i sin nu) when n = 2

    I am so frustrated at this point. I have to list all my steps and no matter how I try to fix it, it never comes out correctly.

    Please help.
    What you are being asked to prove is:

    z^2=r^2(\cos(2u)+i\sin(2u))

    Well:

    z^2=r^2(\cos(u)+i\sin(u))^2=r^2((\cos(u))^2-(\sin(u))^2+2i\sin(u)\cos(u))

    but we have identities:

    \cos(2u)=(\cos(u))^2-(\sin(u))^2

    and

     <br />
\sin(2u)=2\sin(u)\cos(u)<br />

    Hence:

    z^2=r^2(\cos(2u)+i\sin(2u))

    as required

    RonL
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  3. #3
    MHF Contributor kalagota's Avatar
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    In general, it is true for any integer n.

    note that you can also write z = r(\cos\theta + i\sin\theta) as z=re^{i\theta}

    thus, if n \in \mathbb{Z} we have

    z^n=\left(re^{i\theta}\right)^n = r^ne^{i(n\theta)}

    then transform the RHS to trigonometric form, you will get

    z^n = r^n (\cos (n\theta) + i\sin (n\theta))
    Last edited by kalagota; August 1st 2008 at 07:00 PM.
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by kalagota View Post
    In general, it is true for any integer n.

    note that you can also write z = r(\cos\theta i\sin\theta) as z=re^{i\theta}

    thus, if n \in \mathbb{Z} we have

    z^n=\left(re^{i\theta}\right)^n = r^ne^{i(n\theta)}

    then transform the RHS to trigonometric form, you will get

    z^n = r^n (\cos (n\theta) + i\sin (n\theta))
    Once you can write a complex number in exponential form there is nothing left to prove!

    RonL
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