# non-existence of a polynomial [duplicate]

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.

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

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.
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.