analysiscomplex numberscomplex-analysispolynomials
Let $\sum_{k=0}^{n}a_{k}z^{k}$ be a polynomial of degree $n$ with real
coefficients satisfying $$a_{0}>a_{1}>….>a_{n-1}>a_{n}>0$$ Prove
that $p(z)=0$ implies $\left|z\right|>1$.
I have seen similar questions here but none of them proves that solutions can't exist on the unit circle.
Links for similar questions:
Showing that the roots of a polynomial with descending positive coefficients lie in the unit disc.
The thing to do is to look instead at the polynomial $$Q(z) = (1-z)P(z) = (1-z)\left(\sum_{i=0}^n a_iz^i \right) = a_0 -a_n z^{n+1} + \sum_{i=1}^n (a_i-a_{i-1})z^i$$ Now, let $|z|>1$ be a root of $P(z)$, and hence a root of $Q(z)$. Therefore, we have $a_0 + \sum_{i=1}^n (a_i-a_{i-1})z^i = a_n z^{n+1}$ Then, we have \begin{aligned} |a_n z^{n+1}| &= \left|a_0 + \sum_{i=1}^n (a_i-a_{i-1})z^i\right| \\ & \le a_0 + \sum_{i=1}^n (a_i-a_{i-1})|z^i| \\ & < a_0|z^n| + \sum_{i=1}^n (a_i-a_{i-1})|z^n| \\ & = |a_n z^n|\end{aligned} a contradiction.
For a nice article on integer polynomials, see here. (Your problem is Proposition 10)
This is not a answer. I just want to use the space and easy-to-edit feature in the "Answer" section to type the equations.
You did not specify the relation of the modulus of the coefficients $a_n$. If they happen to be cup-shaped, then you may use a theorem by Chen (J. of Math. Anal. and Appl. vol 190, 714-724 (1995)).
By cup-shaped, I mean
$$|a_0|\ge |a_1| \ge \cdots \le |a_{N-2}| \le |a_{N-1}|$$
In your case, we have (assume $N=2n+2$) $$P(z) = \sum_{k=0}^{2n+1}a_kz^k=z^n q(z)+q^*(z)$$
where $$q_n(z)=\sum_{k=0}^na_{n+k+1}z^k$$ $$q_n^*(z)=\sum_{k=0}^n a_kz^k$$
If $N=2n+1$, then we define $Q(z)=(1+z)P(z)$ and treat $Q(z)$ in a similar fashion.
Best Answer
The proof from Let $a_n$ be a decreasing sequence. Prove that the power series $\sum a_n x^n$ has no roots in $A=\{z\in C:|z|<1\}$ can be modified slightly to show that $|z| \le 1$ is not possible for a zero of $p$:
If $p(z) = 0$ and $|z| \le 1$ then $$ \begin{align} a_0 &= \lvert a_0 - (1-z)p(z) \rvert = \left\lvert \sum_{k = 1}^{n} (a_{k-1} - a_k)z^k + a_n z^{n+1} \right\rvert \\ &\underset{(*)}{\le} \sum_{k = 1}^{n} (a_{k-1} - a_k) |z^k| + a_n |z| ^{n+1} \\ &\underset{(**)}{\le} \sum_{k = 1}^{n} (a_{k-1} - a_k) + a_n = a_0 \, . \end{align} $$ So equality holds
The positive factors do not affect the argument, therefore $z, z^2, \ldots, z^n$ must all have the same argument. In particular, $z = z^2/z$ is real and positive, with absolute value $1$, so that $z=1$. But $p(1) \ne 0$, obviously.