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Math Help - Complex number problem!

  1. #1
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    Complex number problem!

    Find Re(z) and Im(z) if:

     z=\sqrt[3]{ \frac{2}{-\sqrt{2+i\sqrt{2}}}}

    My idea was to simplify the expression to the normal form x+iy (The problem is that i don't know how). And then, by De Moivre's formula I can find the roots.

    Any idea on how to simplify the formula and reach the normal form?

    Thank you!
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  2. #2
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    I think you might do better to use the "polar form", r e^{i\theta}= r(cos(\theta)+ i sin(\theta)). That way  z^3= r^3 e^{3i\theta}= \frac{2}{-\sqrt{\frac{2}}{-\sqrt{2+ i\sqrt{2}}}

    z^6= r^6e^{6i\theta}= \frac{4}{2+ i\sqrt{2}} = \frac{4}{2+ i\sqrt{2}}\frac{2- i\sqrt{2}}{2- i\sqrt{2}} = \frac{8- 4i\sqrt{2}}{6}= \frac{4}{3}- \frac{2}{3}\sqrt{2}.

    Put that last number into polar form: its magnitude is \sqrt{\frac{16}{9}+ \frac{4}18}= \sqrt{\frac{36}{18} = \frac{6}{3\sqrt{2}}= \sqrt{2} and its argument is [tex]arctan\left(-\frac{6\sqrt{2}}{12}\right)= arctan\left(-\sqrt{2}{2}}\right) = -\frac{\pi}{4}.

    Now solve r^6= \sqrt{2} and 6\theta= -\frac{\pi}{4}. The real and imaginary parts are r cos(\theta) and r sin(\theta), respectively.
    Last edited by HallsofIvy; August 13th 2010 at 02:56 AM.
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  3. #3
    Super Member 11rdc11's Avatar
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    Should it not be \sqrt{\frac{16}{9} + \frac{8}{9}}= \sqrt{\frac{24}{9}}=\frac{2\sqrt{6}}{3}
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  4. #4
    Senior Member yeKciM's Avatar
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    Quote Originally Posted by 11rdc11 View Post
    Should it not be \sqrt{\frac{16}{9} + \frac{8}{9}}= \sqrt{\frac{24}{9}}=\frac{2\sqrt{6}}{3}
    yes it should, that's just typo

    \displaystyle  |z| == r == \rho == \sqrt{(\Re^2+\Im^2)}

    \displaystyle  \theta == \varphi == \tan^{-1} \frac {\Im}{\Re}
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