1
$\begingroup$

This question already has an answer here:

This is a problem from a practice prelim exam I found online.

Show that there cannot exist a polynomial of the form $p(x) = z^n +a_{n-1}z^{n-1} + \dots + a_1 z + a_0$ such that $|p(z)| < 1$ for all $z$ such that $|z| = 1$.

Here is what I know so far. By Rouche's theorem, if $f(z) = a_{n-1}z^{n-1} + \dots + a_1 z + a_0$ then $f(z)$ has n zeros in the unit circle

If you plug in $z=1$ you get that the sum of $a_i$ has to be less than zero and more than negative two (I think).

I was told to consider $p(1/z)$ but I can't figure out how that helps.

Also perhaps we could use the fact that $p'(0) = a_1$

Also: I am not sure how to show that $p$ does not have a root on the unit circle...

Note: this question was in a section on "Rouche's theorem" in case that helps.

$\endgroup$

marked as duplicate by Martin R, clathratus, mihaild, Shogun, YuiTo Cheng Jun 7 at 1:25

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

2
$\begingroup$

Note that aserting that $\lvert z\rvert=1\implies\bigl\lvert p(z)\bigr\rvert<1$ is equivalent to $\lvert z\rvert=1\implies\bigl\lvert-p(z)\bigr\rvert<\lvert z^n\rvert$. It follows from this that $z^n$ and $-p(z)+z^n$ have the same number of zeros (counted with their multiplicities) in the open unit disk. But $z^n$ has $n$ zeros there whereas$$-p(z)+z^n(=-a_{n-1}z^{n-1}-\cdots-a_1z-a_0)$$has at most $n-1$ zeros there.

$\endgroup$
  • $\begingroup$ Of course. I should remember to consider all forms Rouche's theorem can take. $\endgroup$ – pictorexcrucia Jun 6 at 21:14
1
$\begingroup$

The family of exponential functions $e^{int}$ form an orthonormal system for the scalar product $\displaystyle \langle f,g\rangle=\frac 1{2\pi}\int_0^{2\pi} f\bar g$

So since $|z|=1$ then $z=e^{it}$ and $\displaystyle ||p||^2=\frac 1{2\pi}\int_0^{2\pi} |p(e^{it})|^2\mathop{dt}=\sum\limits_{i=0}^n |a_i|^2$

But $\begin{cases}|p(z)|<1&\implies||p||^2<1\\ a_n=1&\implies\sum\limits_{i=0}^n |a_i|^2\ge 1\end{cases}\quad$ contradiction.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.