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Math Help - De Moivre's Theorem and quadratics

  1. #1
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    De Moivre's Theorem and quadratics

    a) Find all solutions of the equation. (Write your answers in the form a + bi. Enter your answers as a comma-separated list.)


    Answers I got for this is -2 + 2i and -2-2i..I assume you just use the quadratic formula and solve..is that right?

    b) Find the indicated power using De Moivre's Theorem.

    Don't know if I did it right, but I got 1024i for this problem..can someone verify?
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  2. #2
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    Quote Originally Posted by Jgirl689 View Post
    a) Find all solutions of the equation. (Write your answers in the form a + bi. Enter your answers as a comma-separated list.)


    Answers I got for this is -2 + 2i and -2-2i..I assume you just use the quadratic formula and solve..is that right?

    b) Find the indicated power using De Moivre's Theorem.

    Don't know if I did it right, but I got 1024i for this problem..can someone verify?
    Code:
    >> syms x;
    >> solve('x^2 + 2*x + 5 = 0',x)
     
    ans =
     
     - 2*i - 1
       2*i - 1
    Code:
    >> (1+ sqrt(-1))^20
    
    ans =
    
           -1024
    These are the correct answers.
    Last edited by Anonymous1; May 3rd 2010 at 12:35 PM.
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  3. #3
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    [quote=Jgirl689;506170]a) Find all solutions of the equation. (Write your answers in the form a + bi. Enter your answers as a comma-separated list.)


    Answers I got for this is -2 + 2i and -2-2i..I assume you just use the quadratic formula and solve..is that right?

    yes, it is

    b) Find the indicated power using De Moivre's Theorem.

    Don't know if I did it right, but I got 1024i for this problem..can someone verify?

    check again!!!!
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  4. #4
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    Quote Originally Posted by rubic View Post
    b) Find the indicated power using De Moivre's Theorem. Don't know if I did it right, but I got 1024i for this problem..
    Not right, but close.

    \left(1+i\right)^{20} = \left\{\sqrt{2}\left[\cos\left(\dfrac{\pi}{4}\right)+i\sin\left(\dfrac{  \pi}{4}\right)\right]\right\}^{20}  = \left(\sqrt{2}\right)^{20}\left\{\cos\left(\dfrac{  20\pi}{4}\right)+i\sin\left(\dfrac{20\pi}{4}\right  )\right\}  = 2^{10}\bigg\{\left(5\pi\right)+i\sin\left(5\pi\rig  ht)\bigg\}  = 2^{10}\bigg\{\cos\left(\pi\right)+i\sin\left(\pi\r  ight)\bigg\} = 2^{10}\left(-1+i(0)\right) = -2^{10}.
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  5. #5
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    Please help me to find out,A problem on De Moivre`s theorem

    First i apologize if i have posted at wrong place....

    The problem is here:

    Solve x^8+x^5+x^3+1 by using De Moivre`s theorem

    An example in my book might help(didnt helped me at at all)

    x^4-x^3+x^2-x+1

    Sol:

    The equation can be written as

    x^4-x^3+x^2-x+1 = (x^5+1)/(x+1)
    Hence the required roots of x^5+1 = 0 are same as those of (x^4-x^3+x^2-x+1)(x^5+1)

    The equation (x^5+1) gives x^5 = -1 or x = (-1)^1/5

    (......and so on)


    Now what my trouble is how i can find equation like (x^5+1) for my problem stated at start of post?


    I am newbie and poor in maths so little more explanation will help lot
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  6. #6
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    Solve x^8+x^5+x^3+1 by using De Moivre`s theorem.

    The given problem can be factorized as

    x^5(x^3 + 1) + 1(x^3 + 1) = 0

    (x^3+1)(x^5+1) = 0

    Now solve.
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  7. #7
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    thanks
    sa-ri-ga-ma

    I did it....
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