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Thread: complex polynom

  1. #1
    Junior Member
    May 2009

    complex polynom

    hello, i have to determine the complex zero values of the function z^5+2z^3+z

    ok here my try:

    z (z^4+2z^2+1) = 0 => gives first zero pt. at 0
    now substitution z^4 =u^2 and z^2 = u

    now with pq formula i get this: -1+- 0

    but what now ??
    thanks in advance !
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  2. #2
    Junior Member nimon's Avatar
    Sep 2009
    Edinburgh, UK

    Repeated Roots

    Let $\displaystyle z^{5}+2z^{3}+z = p(z).$

    By the remainder theorem you know that:
    $\displaystyle a$ is a root of $\displaystyle p(z) \Leftrightarrow p(z)=(z-a)q(z)$ for some poloynomial $\displaystyle q(z).$
    You did this correctly for $\displaystyle a=0$ to get
    $\displaystyle p(z) = (z-0)q(z) = z(z^{4}+2z^{2}+1).$
    Then by your substitution you found that $\displaystyle u=z^{2}=-1$, hence $\displaystyle z=(-1)^{\frac{1}{2}}=\pm i$, so that we get
    $\displaystyle p(z)=z(z^{2}+1)(z^{2}+1) = z(z+i)(z-i)(z+i)(z-i)$.
    Therefore the roots are $\displaystyle 1,i,-i$. The reason that you got 3 distinct roots instead of the 5 you might expect with an order 5 polynomial is that the factors $\displaystyle (z+i)$ and $\displaystyle (z-i)$ are both repeated in the (unique) factorisation of $\displaystyle p(z).$ So really there are 5 roots, it's just that $\displaystyle i$ and $\displaystyle -i$ are repeated.
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