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Math Help - please help with this

  1. #1
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    please help with this

    Find the realand complex solutions to the equation X3(cubed) + 1=0

    my choices

    a. -1,1,cis(pi/3)
    b. 1, cis(pi/3), cis (2pi/3)
    c. -1, cis(pi/3), cis(5pi/3)
    d. cis (pi/3), cis(2pi/3), cis(4pi/3)
    Last edited by ccdalamp; July 10th 2007 at 12:38 PM.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by ccdalamp View Post
    Find the realand complex solutions to the equation X3(cubed) + 1=0

    my choices

    a. -1,1,cis(pi/3)
    b. 1, cis(pi/3), cis (2pi/3)
    c. -1, cis(pi/3), cis(5pi/3)
    d. cis (pi/3), cis(2pi/3), cis(4pi/3)
    x^3=-1= cis((2n+1) \pi )\ \ \ n=0,1,2, ...

    so

    <br />
x=cis\left(\frac{2n+1}{3}\pi \right)\ \ \ n=0,1,2, ...<br />

    which if I am not mistaken is answer (c)

    RonL
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  3. #3
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    Hello, ccdalamp!

    With "cis" in the answer choices, I assume that DeMoivre's Formula is expected.
    But it can be solved algebraically . . . with some trig.


    Find the real and complex solutions to the equation: . x^3 + 1\:=\:0

    a)\;\text{-}1,\:1,\:\text{cis}\left(\frac{\pi}{3}\right)\qqua  d\qquad\qquad b)\;1,\:\text{cis}\left(\frac{\pi}{3}\right),\:\te  xt{cis}\left(\frac{2\pi}{3}\right)
    c)\;\text{-}1,\:\text{cis}\left(\frac{\pi}{3}\right),\:\text{  cis}\left(\frac{5\pi}{3}\right)\qquad d)\;\text{cis}\left(\frac{\pi}{3}\right),\:\text{c  is}\left(\frac{2\pi}{3}\right),\:\text{cis}\left(\  frac{4\pi}{3}\right)

    Factor: . (x + 1)(x^2 - x + 1) \:=\:0

    And we have: . \begin{array}{ccccccc}x + 1 & = & 0 & \;\;\Rightarrow\;\; & x & = & -1 \\<br />
x^2-x+1 & = & 0 & \Rightarrow & x & = & \frac{1\pm i\sqrt{3}}{2} \end{array}

    And the complex roots can be rewritten: . \begin{array}{ccc}\frac{1}{2} + \frac{\sqrt{3}}{2}i & \Rightarrow & \cos\left(\frac{\pi}{3}\right) + i\sin\left(\frac{\pi}{3}\right) \\<br />
\frac{1}{2} - \frac{\sqrt{3}}{2}i & \Rightarrow & \cos\left(\frac{5\pi}{3}\right) + i\sin\left(\frac{5\pi}{3}\right) \end{array}

    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    On the other hand, there is an "eyeball" solution.

    A cubic equation has three roots, of course.
    Either all three are real or there is one real and two complex conjugates.
    . . Only (c) meets this requirement.

    Moreover, we already know that: . \sqrt[3]{\text{-}1} = \text{-}1

    Therefore, (c) is the only candidate.

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