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Math Help - Solving an equation with de Moivre's formula

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

    z^2=-119+120i


    z=r(cosv+isinv)

    and

     <br />
-119+120i=(-119+120i)(cos0+isin0)<br />


    De Moivres states
     <br />
z^2=r^2(cos2v+isin2v)<br />

    which leads to

    z^2=-119+120i

    can be written

     <br />
(-119+120i)(cos0+isin0)=r^2(cos2v+isin2v)<br />


    The two lines equals eachother if

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



    This is my problem:

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

    Joined
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    From
    Lexington, MA (USA)
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    Hello, Greenb!

    Is this the problem?


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


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

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


    Using half-angle identities:

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

    . . \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]: . 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;

    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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