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Thread: Sqaure roots of complex number

  1. #1
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    Sqaure roots of complex number

    Hi,

    The question i have is "Find the sqaure roots of $\displaystyle \sqrt{3} + i$ "

    The book gives the answer:
    $\displaystyle \sqrt{2}e^{(\frac{1}{2} + k)\pi i}$
    $\displaystyle = \pm \sqrt{2}(cos(\frac{\pi}{12}) + isin(\frac{\pi}{12}))$

    I can see how they get that but I thought you have to plug in k = 0, 1(n-1) and then use the the new arguements given from that to find them.

    I get $\displaystyle arg = \frac{\pi}{6}, \frac{7\pi}{6}$

    Which leads my answer to be:

    $\displaystyle \pm(\frac{\sqrt{6}}{2} + i\frac{\sqrt{2}}{2})$


    Could anyone clear up my grave error?

    Many thanks
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  2. #2
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    Maybe

    Im not sure of this will work but use Moivre´s.
    Wright the complex number in polar form and set that to be $\displaystyle z^2$
    Then solve for Z....
    Hope it was helpful..
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  3. #3
    MHF Contributor Calculus26's Avatar
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    Simply factor out a sqrt(2) from

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  4. #4
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    Turns out like this for me

    :$\displaystyle sqrt(3) +i$
    $\displaystyle abs:z = 2$
    $\displaystyle argz = atan1/sqrt3) = 30deg$

    $\displaystyle => 2(cos(pi/6)+isin(pi/6)$
    Set that to be $\displaystyle z^2$

    $\displaystyle z = sqrt(2)(cos(pi/12) +isin(pi/12))$ + the other root that gave the +/- sign to the equation...
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  5. #5
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    Thanks for the tips,

    When I take out the factor of root 2 I have :

    $\displaystyle
    \pm \sqrt{2}(cos(\frac{\pi}{6}) +isin(\frac{\pi}{6}))
    $

    Where apparently im supposed to have $\displaystyle \theta = \frac{\pi}{12}$
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  6. #6
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    Divide the argument

    Need to devide the argument (pi/6) by 2. $\displaystyle Z^n$ has $\displaystyle n$ roots, the arguments of the roots are evently spread around a circle.

    To find roots of $\displaystyle z^2$ devide the argument by $\displaystyle 2 => (pi/6)2$ and that gives you pi/12.... the other roots are found by:

    $\displaystyle
    (2*pi*n)/2 = pi*n
    $ where your n can be 1;2
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  7. #7
    MHF Contributor Calculus26's Avatar
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    Perhaps the following attachment may help
    Attached Thumbnails Attached Thumbnails Sqaure roots of complex number-complex.jpg  
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  8. #8
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    Great, thanks for all the help.

    As a side note could anyone recommend a book/website with exercises to do with complex number? I have 2 already but they lack exercises.

    Thanks
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