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    Polynomial Equations : Solutions, Conjugates and Shift Operator

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    Prove that if p is a polynomial with real coefficients, and if is a (complex) solution of P(E)z = 0, then the conjugate of z, the real part of z, and the imaginary part of z are also solutions.

    Note: This is from a numerical analysis course, and here P(E) refers to a polynomial in E, the "shift operator" for a sequence.

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    https://brainmass.com/math/complex-analysis/polynomial-equations-solutions-conjugates-shift-operator-49807

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    Hi,
    _
    If z is a complex number, and z its conjugate, first you need to prove that conjugation distributes over multiplication and addition:
    ______ _ _
    (z1 * z2) = z1 * z2
    _______ _ _
    (z1 + z2) = (z1) + (z2)

    If z= x+iy, then the complex number x-iy is called the conjugate of the complext number z and is
    _ __ __
    written as z . It is easily seen that numbers conjugate to z1+z2 and z1z2 are (z1) + (z2) and
    _ _
    z1 z2 respectively.

    Let P(x) = a x^n + a x^n-1 + a x^n-2 +...+ a x + a
    ...

    Solution Summary

    Polynomials, complex solutions, conjugates and shift operators are investigated.

    $2.49

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