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Thread: complex number question

  1. #1
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    complex number question

    $\displaystyle Z = \frac{8(1 + i)}{\sqrt{2}} $

    write Z as r(cosΘ +i sinΘ)[/tex]

    using that show the three values for $\displaystyle Z^{2/3}$ are

    $\displaystyle -4i , 2(\sqrt{3} + i ) ,2(-3\sqrt{3} + i ) $


    ----
    i can do the first part

    $\displaystyle z=8(cos(\pi/4) + i sin(\pi/4)) $

    and since $\displaystyle z^n =8^n (cos( n* \pi/4) + i sin(n*\pi/4))$

    $\displaystyle
    z^{2/3} = 4( cos\pi/6 + isin\pi/6)=2(\sqrt{3} + i)$

    i cant get other 2 numbers
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  2. #2
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    Quote Originally Posted by stud_02 View Post
    $\displaystyle Z = \frac{8(1 + i)}{\sqrt{2}} $

    write Z as r(cosΘ +i sinΘ)[/tex]

    using that show the three values for $\displaystyle Z^{2/3}$ are

    $\displaystyle -4i , 2(\sqrt{3} + i ) ,2(-3\sqrt{3} + i ) $


    ----
    i can do the first part

    $\displaystyle z=8(cos(\pi/4) + i sin(\pi/4)) $

    and since $\displaystyle z^n =8^n (cos( n* \pi/4) + i sin(n*\pi/4))$

    $\displaystyle
    z^{2/3} = 4( cos\pi/6 + isin\pi/6)=2(\sqrt{3} + i)$

    i cant get other 2 numbers
    Since sine and cosine are periodic with period $\displaystyle 2\pi$, z is also equal to $\displaystyle z= 8(cos(\pi/4+ 2\pi)+ i sin(\pi/4+ 2\pi))$ and $\displaystyle z= 8(cos(\pi/4+ 4\pi)+ i sin(\pi/4+ 4\pi))$.

    Apply that same formula to this.
    Last edited by HallsofIvy; Jul 28th 2009 at 04:21 AM.
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