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Math Help - How do I find z using the theorem of De Moivre

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    How do I find z using the theorem of De Moivre

    I was going through some old work sheets and couldn't seem to do this one for some reason...

    If z^2=2-2i find z using the theorem of De Moivre.

    What I did:

    z=\sqrt{2-2i}

    8^{\frac{1}{4}} (cis\left(\frac{7\pi}{8}\right))

    I assume I would find values for cos\left(\frac{7\pi}{8}\right) and sin\left(\frac{7\pi}{8}\right) and then I would be done right? Assuming this is what I am supposed to do, how do I solve for the angle \frac{7\pi}{8}?

    Thanks.
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    Re: How do I find z using the theorem of De Moivre

    Quote Originally Posted by terrorsquid View Post
    I was going through some old work sheets and couldn't seem to do this one for some reason...
    If z^2=2-2i find z using the theorem of De Moivre.
    Let z^2=2-2i=\sqrt{8}\exp \left( {\frac{{ - \pi i}}{4}} \right)
    so z=\omega = \sqrt[4]{8}\exp \left( {\frac{{ - \pi i}}{8}} \right)

    If \varsigma  = \exp \left( {\pi i} \right) then the two roots are \omega\cdot\varsigma^k,~k=0,1~.
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    Re: How do I find z using the theorem of De Moivre

    Quote Originally Posted by terrorsquid View Post
    I was going through some old work sheets and couldn't seem to do this one for some reason...

    If z^2=2-2i find z using the theorem of De Moivre.

    What I did:

    z=\sqrt{2-2i}

    8^{\frac{1}{4}} (cis\left(\frac{7\pi}{8}\right))

    I assume I would find values for cos\left(\frac{7\pi}{8}\right) and sin\left(\frac{7\pi}{8}\right) and then I would be done right? Assuming this is what I am supposed to do, how do I solve for the angle \frac{7\pi}{8}?

    Thanks.
    \displaystyle |2 - 2i| = \sqrt{2^2 + (-2)^2} = \sqrt{8} = 2\sqrt{2} and \displaystyle \arg{(2-2i)} = -\arctan{\frac{2}{2}} = -\frac{\pi}{4}.

    So \displaystyle 2 - 2i = 2\sqrt{2}\,\textrm{cis}\,\left(-\frac{\pi}{4}\right).


    You have

    \displaystyle \begin{align*} z^2 &= 2\sqrt{2}\,\textrm{cis}\left(-\frac{\pi}{4}\right) \\ z &= \left[2\sqrt{2}\,\textrm{cis}\left(-\frac{\pi}{4}\right)\right]^{\frac{1}{2}} \\ &= \sqrt{2\sqrt{2}}\,\textrm{cis}\left[\frac{1}{2}\left(-\frac{\pi}{4}\right)\right] \\ &= \sqrt[4]{8}\,\textrm{cis}\left(-\frac{\pi}{8}\right) \\ &= \sqrt[4]{8}\left[\cos{\left(-\frac{\pi}{8}\right)} + i\sin{\left(-\frac{\pi}{8}\right)}\right] \\ &= \sqrt[4]{8}\left[\cos{\left(\frac{\pi}{8}\right)} - i\sin{\left(\frac{\pi}{8}\right)}\right]\end{align*}

    You will need to use half angle identities to evaluate these exactly.

    There is also another square root. You should know that square roots are evenly spaced about a circle, so these two square roots will have the same magnitude and are separated by an angle of \displaystyle \pi, so the other square root is

    \displaystyle \begin{align*} \sqrt[4]{8}\,\textrm{cis}\left(\pi - \frac{\pi}{8}\right) &= \sqrt[4]{8}\left[\cos{\left(\pi - \frac{\pi}{8}\right)} + i\sin{\left(\pi - \frac{\pi}{8}\right)}\right] \\ &= \sqrt[4]{8} \left[ -\cos{ \left(\frac{\pi}{8}\right) } + i \sin{  \left (  \frac{\pi}{8}\right) }  \right]\end{align*}

    and you should have those values from the first square root
    Last edited by Prove It; October 1st 2011 at 07:17 AM.
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