# Complex Polynomial With Complex Coefficients

• Jun 27th 2011, 11:02 PM
Oipiah
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.)
• Jun 28th 2011, 12:49 AM
ojones
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?
• Jun 28th 2011, 08:41 AM
Oipiah
Re: Complex Polynomial With Complex Coefficients
No I am not
• Jun 28th 2011, 11:46 AM
Opalg
Re: Complex Polynomial With Complex Coefficients
Quote:

Originally Posted by Oipiah
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.]
• Jun 28th 2011, 12:53 PM
hatsoff
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 $a_nz^n$, 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.
• Jun 28th 2011, 02:04 PM
Opalg
Re: Complex Polynomial With Complex Coefficients
Quote:

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 $a_nz^n$, 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 $\pi.$
• Jun 28th 2011, 04:59 PM
ojones
Re: Complex Polynomial With Complex Coefficients
Quote:

Originally Posted by Oipiah
No I am not

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