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Thread: Solving an equation with de Moivre's formula

  1. #1
    Junior Member
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    Solving an equation with de Moivre's formula

    $\displaystyle z^2=-119+120i$


    $\displaystyle z=r(cosv+isinv)$

    and

    $\displaystyle
    -119+120i=(-119+120i)(cos0+isin0)
    $


    De Moivres states
    $\displaystyle
    z^2=r^2(cos2v+isin2v)
    $

    which leads to

    $\displaystyle z^2=-119+120i$

    can be written

    $\displaystyle
    (-119+120i)(cos0+isin0)=r^2(cos2v+isin2v)
    $


    The two lines equals eachother if

    $\displaystyle z^2=-119+120i$
    and
    $\displaystyle v=0+n*pi$
    making the angles
    $\displaystyle v1=0$
    $\displaystyle v2=pi$



    This is my problem:

    $\displaystyle z^2=-119+120i$
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  2. #2
    Super Member

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    Lexington, MA (USA)
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    Hello, Greenb!

    Is this the problem?


    $\displaystyle z^2\:=\:-119+120i$. .Find $\displaystyle z.$
    On the argand diagram, we have:
    Code:
            *
            : * 169
        120 :   * 
            :     * θ
        - - + - - - * - - - -
              -119
    Hence: .$\displaystyle \begin{array}{ccc}\cos\theta &=&\text{-}\frac{119}{169} \\ \\[-4mm] \sin\theta &=& \frac{120}{169} \end{array}$


    We have: .$\displaystyle z^2 \;=\;169(\cos\theta + i\sin\theta)$

    And we want: .$\displaystyle z \;=\;13\left(\cos\tfrac{\theta}{2} + i\sin\tfrac{\theta}{2}\right)$ .[1]


    Using half-angle identities:

    . . $\displaystyle \cos\tfrac{\theta}{2} \:=\:\sqrt{\frac{1+\cos\theta}{2}} \:=\:\sqrt{\frac{1-\frac{119}{169}}{2}} \;=\;\sqrt{\frac{50}{338}} \:=\:\frac{5}{13}$

    . . $\displaystyle \sin\tfrac{\theta}{2} \:=\:\sqrt{\frac{1-\cos\theta}{2}} \:=\:\sqrt{\frac{1-(\text{-}\frac{119}{169})}{2}} \;=\;\sqrt{\frac{288}{338}} \;=\;\frac{12}{13}$


    Substitute into [1]: .$\displaystyle z \;=\;13\left(\frac{5}{13} + i\frac{12}{13}\right) \quad\Rightarrow\quad\boxed{ z\;=\;5 + 12i}$

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  3. #3
    Junior Member
    Joined
    Jan 2009
    Posts
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    Ah I forgot that |z| equals the hypotenuse, but I'm not with you all the way on the half angle identities part, seeing that its a cubicequation, it should have two roots.


    And what if the equation was;

    $\displaystyle z^4=4-4i$

    could it be solved the same way?

    I'm Swedish by the way so sorry if I don't get the termonoligy right.
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