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.)

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?

Re: Complex Polynomial With Complex Coefficients

Re: Complex Polynomial With Complex Coefficients

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Originally Posted by

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

Let $\displaystyle p(z) = a_nz^n+\ldots+a_1z+a_0$ be a polynomial of degree n with complex coefficients and suppose there exist $\displaystyle \{z_1,\ldots,z_n\}$ distinct complex numbers each having $\displaystyle \text{Im}(z_j)>0$ such that $\displaystyle p(z_j)=0$ for $\displaystyle j\in\{1,\ldots,n\}.$ Set $\displaystyle q(z)=\text{Re}(a_n)z^n+\ldots+\text{Re}(a_1)z+ \text{Re}(a_0).$ Prove that if $\displaystyle q(w) = 0$ then w must be real.

(Here, $\displaystyle \text{Im}(z_j)$ denotes the imaginary part of $\displaystyle z_j$ and $\displaystyle \text{Re}(a_j)$ denotes the real part of $\displaystyle a_j$.)

The key to this is to think in terms of the winding number. Let $\displaystyle D$ 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 $\displaystyle D$. It follows that as z goes round $\displaystyle D$, p(z) will go n times round the origin (in other words, its winding number is n). Thus $\displaystyle \text{Re}(p(z))$ 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 $\displaystyle a_nz^n$, and so its real part will take the value 0 n times. Therefore the other n zeros of $\displaystyle \text{Re}(p(z))$ 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 $\displaystyle \text{Re}(p(z)) = q(z)$ if z is real, but not otherwise.]

Re: Complex Polynomial With Complex Coefficients

Opalg,

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

Quote:

Originally Posted by

**Opalg** As z goes round the semicircle (so that |z|=R, which is large), p(z) will be dominated by its first term $\displaystyle a_nz^n$, 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.

Re: Complex Polynomial With Complex Coefficients

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Originally Posted by

**hatsoff** Quote:

Originally Posted by

**Opalg** As z goes round the semicircle (so that |z|=R, which is large), p(z) will be dominated by its first term $\displaystyle a_nz^n$, 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 $\displaystyle a_nz^n.$ If you put $\displaystyle z = Re^{i\theta} \ (0\leqslant\theta\leqslant\pi)$, then the real part of $\displaystyle z^n$ is $\displaystyle \cos(n\theta)$, which changes sign n times as $\displaystyle \theta$ goes from 0 to $\displaystyle \pi.$

Re: Complex Polynomial With Complex Coefficients

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Originally Posted by

**Oipiah** No I am not

Did this question come up in the context of control theory or complex analysis?