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Thread: Complex numbers-proving and finding roots

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    Complex numbers-proving and finding roots

    Prove that 1+i is a root of the equation z^4+3z^2-6z+10=0. Find all the other roots.
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    Quote Originally Posted by kingkaisai2
    Prove that 1+i is a root of the equation z^4+3z^2-6z+10=0. Find all the other roots.
    $\displaystyle z^4+3z^2-6z+10=0$

    $\displaystyle (1+i)^4+3(1+i)^2-6(1+i)+10$

    $\displaystyle (1^4+4*1^3*i+6*1^2*i^2+4*1*i^3+i^4)$$\displaystyle +3(1^2+2*1*i+i^2)-6(1+i)+10$

    $\displaystyle (1 + 4i - 6 - 4i + 1)+ 3(1 + 2i - 1) - 6(1 + i) + 10$

    $\displaystyle -4 + 6i - 6 - 6i + 10$

    $\displaystyle 0$ (Check.)

    Now, if 1+i is a root, so is 1-i. Thus we know that $\displaystyle \left ( z - (1+i) \right ) \left ( z - (1-i) \right )$$\displaystyle = \left ( (z-1) - i \right ) \left ( (z-1) + i \right )$$\displaystyle = (z-1)^2 - i^2 = z^2 - 2z + 1 + 1 = z^2 - 2z + 2$
    is a factor of $\displaystyle z^4+3z^2-6z+10$. So do the division. This gives you a quadratic in z to solve, which you can do by factoring or the quadratic formula or whatnot.

    -Dan
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