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Math Help - Polynomial with real coefficients but complex roots

  1. #1
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    Polynomial with real coefficients but complex roots

    Hi everyone

    I'm trying to solve this equation:

    z4+1 = 0

    I've tried every method I can think of, such as De Moivre's formula, but I only get two complex roots.


    What method do I use ?

    I just want to know the method/procedure.
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  2. #2
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    Re: Polynomial with real coefficients but complex roots

    You could rewrite the equation as:

    z^4-i^2=0
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    Re: Polynomial with real coefficients but complex roots

    Should I still use De Moivre's formula ? Or ?
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    Re: Polynomial with real coefficients but complex roots

    You could use Euler/de Moivre (which are easier), but another method would be to set:

    z^2=\pm i

    (a+bi)^2=\pm i

    (a^2-b^2)+(2ab)i=0\pm i

    By equating coefficients, you may determine the 4 roots.

    Using Euler and de Moivre:

    e^{4\theta i}=\cos(\pi+2k\pi)+i\sin(\pi+2k\pi)

    \theta=\frac{\pi}{4}(2k+1) where k\in\{0,1,2,3\}

    Can you finish?
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    Re: Polynomial with real coefficients but complex roots

    I have to be honest and say that both methods seem a bit unclear to me.
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    Re: Polynomial with real coefficients but complex roots

    z=e^{\theta i}=\cos(\theta)+i\sin(\theta)

    Now, use the 4 values we found above for \theta, i.e., \theta=\frac{\pi}{4},\frac{3\pi}{4},\frac{5\pi}{4}  ,\frac{7\pi}{4}.
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    Re: Polynomial with real coefficients but complex roots

    Can you explain step 1 in Euler and De Moivre ?
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  8. #8
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    Re: Polynomial with real coefficients but complex roots

    We have:

    z^4=-1=-1+0i

    We find: |z|=\sqrt{(-1)^2+0^2}=1 hence, we may state:

    z=e^{\theta i}\:\therefore\:z^4=e^{4\theta i}

    So, we must have:

    e^{4\theta i}=-1+0i=\cos(\pi+2k\pi)+i\sin(\pi+2k\pi) where k\in\mathbb{Z}

    By Euler's formula, we know then:

    4\theta=\pi(2k+1)

    \theta=\frac{\pi}{4}(2k+1)

    Now, we want:

    0\le \theta<2\pi

    0\le \frac{\pi}{4}(2k+1)<2\pi

    0\le 2k+1<8

    Since k is an integer, we have k\in\{0,1,2,3\}

    Hence, we have:

    z=\cos\left(\frac{\pi}{4}(2k+1) \right)+i\sin\left(\frac{\pi}{4}(2k+1) \right)

    Now, use the values we found for k to compute the 4 roots.
    Thanks from Tala
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  9. #9
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    Re: Polynomial with real coefficients but complex roots

    I understand now !

    Thank you so much
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    Re: Polynomial with real coefficients but complex roots

    Hello, Tala!

    \text{Solve: }\:z^4 +1 \:=\: 0

    DeMoivre's Theorem is the approach I would use.


    We have: . z^4 \;=\;-1 \;=\;\cos(\pi\!+\!2\pi n) + i\sin(\pi\!+\!2\pi n)

    Hence: . . z \;=\;\bigg[\cos(\pi\!+\!2\pi n) + i\sin(\pi\!+\!2\pi n)\bigg]^{\frac{1}{4}}

    n . . . . . . z\;=\; \cos\left(\tfrac{\pi}{4}\!+\!\tfrac{\pi}{2}n\right  ) + i\sin\left(\tfrac{\pi}{4}\!+\!\tfrac{\pi}{2}n \right) \quad \text{ for }n = 0,1,2,3
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