Results 1 to 3 of 3

Thread: please help with this

  1. #1
    Newbie
    Joined
    Jul 2007
    Posts
    8

    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; Jul 10th 2007 at 12:38 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    5
    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)
    $\displaystyle x^3=-1= cis((2n+1) \pi )\ \ \ n=0,1,2, ...$

    so

    $\displaystyle
    x=cis\left(\frac{2n+1}{3}\pi \right)\ \ \ n=0,1,2, ...
    $

    which if I am not mistaken is answer (c)

    RonL
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    12,028
    Thanks
    848
    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: .$\displaystyle x^3 + 1\:=\:0$

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

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

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

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

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

    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: .$\displaystyle \sqrt[3]{\text{-}1} = \text{-}1$

    Therefore, (c) is the only candidate.

    Follow Math Help Forum on Facebook and Google+


/mathhelpforum @mathhelpforum