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Math Help - Solve 7th degree equation using complex number

  1. #1
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    Exclamation Solve 7th degree equation using complex number

    Solve equations using De Moivre's theorem:

    1. x^7 + x^4 + x^3 + 1 = 0

    2. x^7 - x^4 + x^3 - 1 = 0


    I tried multiplying with (x - 1). Also tried putting x^3 = y; didn't work.
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  2. #2
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    Re: Solve 7th degree equation using complex number

    Hey anil86.

    Have you been able to factorize the expression in terms (ax + b)^n? (Hint: Consider (x+iy)^n where the real terms are x^7, x^4, x^3 and 1).
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  3. #3
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    Re: Solve 7th degree equation using complex number

    Hello, anil86!

    I'll walk you through the first one . . .



    Solve equations using De Moivre's theorem:

    (1)\;x^7 + x^4 + x^3 + 1 \:=\: 0

    Factor: . x^4(x^3+1) + (x^3+1) \:=\:0

    Factor: . (x^3+1)(x^4+1) \:=\:0


    x^3 + 1 \,=\,0 \quad\Rightarrow\quad x^3 \,=\,-1 \quad\Rightarrow\quad x \,=\,\sqrt[3]{\text{-}1}

    -1 \:=\:\cos(\pi + 2\pi n) + i\sin(\pi + 2\pi n)

    (-1)^{\frac{1}{3}} \;=\;\cos\left(\tfrac{\pi}{3} + \tfrac{2\pi}{3}n\right) + i\sin\left(\tfrac{\pi}{3} + \tfrac{2\pi}{3}n\right) \;\text{ for }n = 0,1,2


    x^4 + 1 \,=\,0 \quad\Rightarrow\quad x^4 \,=\,-1 \quad\Rightarrow\quad x \,=\,\sqrt[4]{\text{-}1}

    -1 \:=\:\cos(\pi + 2\pi n) + i\sin(\pi + 2\pi n)

    (-1)^{\frac{1}{4}} \;=\;\cos\left(\tfrac{\pi}{4} + \tfrac{\pi}{2}n\right) + i\sin\left(\tfrac{\pi}{4} + \tfrac{\pi}{2}n\right) \;\text{ for }n = 0,1,2,3
    Thanks from anil86
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