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Math Help - Complex Polynomial With Complex Coefficients

  1. #1
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    Complex Polynomial With Complex Coefficients

    I can't figure out where to start on this one.

    Let p(z) = an*z^n+...+a1*z+a0 be a polynomial of degree n with complex coefficients and suppose there exists {z1,...,zn} distinct complex numbers each having Imaginary(zj)>0 such that p(zj)=0 for j∈{1,...,n}. Set q(z)=Real(an)z^n+...+Real(a1)z+Real(a0). Prove that if q(w) = 0 then w must be real.
    (Here, Imaginary(zj) denotes the imaginary part of zj and Real(aj) denotes the real part of aj.)
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    Re: Complex Polynomial With Complex Coefficients

    I think this is related to the Hermite-Biehler theorem from control theory. Are you familiar with this theorem?
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    Re: Complex Polynomial With Complex Coefficients

    No I am not
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    Re: Complex Polynomial With Complex Coefficients

    Quote Originally Posted by Oipiah View Post
    I can't figure out where to start on this one.

    Let p(z) = a_nz^n+\ldots+a_1z+a_0 be a polynomial of degree n with complex coefficients and suppose there exist \{z_1,\ldots,z_n\} distinct complex numbers each having \text{Im}(z_j)>0 such that p(z_j)=0 for j\in\{1,\ldots,n\}. Set q(z)=\text{Re}(a_n)z^n+\ldots+\text{Re}(a_1)z+ \text{Re}(a_0). Prove that if q(w) = 0 then w must be real.
    (Here, \text{Im}(z_j) denotes the imaginary part of  z_j and \text{Re}(a_j) denotes the real part of a_j.)
    The key to this is to think in terms of the winding number. Let D denote a D-shaped contour consisting of a long interval [R,R] on the real axis together with a semicircle of radius R around the upper half-plane. If R is large enough then all the zeros of p(z) will be inside D. It follows that as z goes round D, p(z) will go n times round the origin (in other words, its winding number is n). Thus \text{Re}(p(z)) will go from positive to negative and back again n times, and will therefore take the value 0 2n times. As z goes round the semicircle (so that |z|=R, which is large), p(z) will be dominated by its first term a_nz^n, and so its real part will take the value 0 n times. Therefore the other n zeros of \text{Re}(p(z)) must occur as z goes along the real axis. Thus q(z) has n real zeros. Since it is a polynomial of degree n, it cannot have any other zeros.

    [Notice that \text{Re}(p(z)) = q(z) if z is real, but not otherwise.]
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    Re: Complex Polynomial With Complex Coefficients

    Opalg,

    Any chance you could do me the favor of explaining this further:

    Quote Originally Posted by Opalg View Post
    As z goes round the semicircle (so that |z|=R, which is large), p(z) will be dominated by its first term a_nz^n, and so its real part will take the value 0 n times.
    ...? I don't see how one follows from the other.

    It's not important, though, so don't bother if it's inconvenient. But I was curious.
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    Re: Complex Polynomial With Complex Coefficients

    Quote Originally Posted by hatsoff View Post
    Quote Originally Posted by Opalg View Post
    As z goes round the semicircle (so that |z|=R, which is large), p(z) will be dominated by its first term a_nz^n, and so its real part will take the value 0 n times.
    ...? I don't see how one follows from the other.
    The idea is that when R is very large, p(z) will be approximately equal to its first term a_nz^n. If you put z = Re^{i\theta} \ (0\leqslant\theta\leqslant\pi), then the real part of z^n is \cos(n\theta), which changes sign n times as \theta goes from 0 to \pi.
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    Re: Complex Polynomial With Complex Coefficients

    Quote Originally Posted by Oipiah View Post
    No I am not
    Did this question come up in the context of control theory or complex analysis?
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