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Exercise 1.8.5 of Berenstein-Gay "Complex variables" asks to count the number of zeroes in $(0,2\pi)$ of certain trigonometric polynomial.

Towards the exercise, the book suggests to first show the following:

Given $0\le a_0<\dots<a_n$, the roots of $$p(z)=a_0+a_1z+\dots+a_nz^n$$ lie in the unit disk.

Do you see how to solve the above?

(The exercise is used in several places later through the book, so I am very frustrated with myself for not being able to do this on my own. The original version of this posting asked for the whole exercise, which is somewhat unreasonable, as it is really several very different questions together. I am editing the question to address directly the part that has been answered. I may post the rest of 1.8.5 at a later date.)

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1 Answer 1

up vote 3 down vote accepted

For the common exercise here are two similar proofs, but one is completely elementary:


Proof 1

This is the elementary proof.

Consider

$$\frac{p(z)(z-1)}{z^{n+1}} = a_n + \sum_{k=0}^{n-1} \frac{a_k - a_{k+1}}{z^{n-k}} - \frac{a_0}{z^{n+1}} = a_n - \sum_{k=0}^{n-1} \frac{a_{k+1} - a_k}{z^{n-k}} - \frac{a_0}{z^{n+1}} $$

Thus $$\displaystyle \left|\frac{p(z)(z-1)}{z^{n+1}}\right| \ge a_n - \sum_{k=0}^{n-1} \frac{a_{k+1} - a_k}{|z^{n-k}|} - \frac{a_0}{|z^{n+1}|}$$

using $\displaystyle |z_1 - z_2| \ge |z_1| - |z_2|$

Now if $\displaystyle |z| \gt 1$, then $\displaystyle -\frac{a_{k+1} - a_k}{|z^{n-k}|} \gt -(a_{k+1} - a_k)$ and $\displaystyle -\frac{a_0}{|z^{n+1}|} \gt - a_0$

Thus

$$\displaystyle \left|\frac{p(z)(z-1)}{z^{n+1}}\right| \gt a_n - \sum_{k=0}^{n-1} (a_{k+1} - a_k) - a_0 = 0$$

Thus if $\displaystyle |z| \gt 1$, then $\displaystyle |p(z)| \neq 0$ and so all the roots of $\displaystyle p(z)$ satisfy $\displaystyle |z| \le 1$.


Proof 2

Similar to the above consider the polynomial

$$g(z) = p(z)(1-z) + a_n z^{n+1} = a_0 + \sum_{k=0}^{n-1} (a_{k+1} - a_k) z^{k+1}$$

Now given any $\displaystyle c > 0$, we have that, for any $\displaystyle z$ on the boundary of the unit circle,

$$|g(z)| \le a_0 + \sum_{k=0}^{n-1} (a_{k+1} - a_k) = a_n \lt |(a_n + c)z^{n+1}|$$

Thus if $\displaystyle f(z) = (a_n + c) z^{n+1}$, we have that $\displaystyle |g(z)| \lt |f(z)|$ on the boundary of the unit disk and thus both $\displaystyle g(z)-f(z)$ and $\displaystyle f(z)$ have the same number of roots satisfying $\displaystyle |z| \lt 1$, by using Rouche's Theorem.

Thus for any $\displaystyle c > 0$, the polynomial $\displaystyle p(z)(1-z) - c z^{n+1}$ has all its roots within the unit circle, i.e. the root $\displaystyle w$ with maximum modulus satisfies $\displaystyle |w| \lt 1$.

Now it is a well known fact that the eigenvalues of a complex matrix are continuous and thus the function of the root with the maximum modulus is continuous (though I am not completely sure of this). Taking limits as $\displaystyle c \to 0$ we get the result that the roots of $\displaystyle p(z)(1-z)$ all satisfy $\displaystyle |z| \le 1$.

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Many thanks, I am thinking about this. Continuity of the root (in your second answer) is true---actually, the book discusses this in section 2.6, in the context of the Residue Theorem. –  Bruce George Jan 20 '11 at 0:03
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