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Suppose that the coefficients of the equation $x^n+a_{n-1}x^{n-1}+\cdots +a_1x+a_0=0$ are real and satisfy $0<a_0 \le a_1 \le \cdots \le a_{n-1} \le 1$.

Let $z$ be a complex root of the equation with $|z| \ge 1$. Show that $z^{n+1}=1$ $\space\space\space\space\space\space\space\space\space\space\space\space$(Source:MOP)

My work

Let $0<a_0=a_1 = \cdots = a_{n-1}=1 $, then, given that $z$ is a complex root of the equation $x^n+a_{n-1}x^{n-1}+\cdots a_1x+a_0=0$, we have that $$z^n+z^{n-1}+\cdots+z+1=0 $$

Now,if we let $n=k-1$ we have $$z^{k-1}+z^{k-2}+\cdots+1=0 $$

From this it follows that $z$ is a root of unity which satisfy the equation $x^k-1=0$ ,so we have that $z^k=1$ where $k=n+1$ so we have $z^{n+1}=1$


That's what I've been able to do but I am still unsatisfied as I have considered a special condition, namely $0<a_0=a_1 = \cdots = a_{n-1}=1 $, and I would like to see how to take care also of the other condition.

How can I do that?

Thanks in advance

Edit: The current answer is just too hard for me to understand.

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  • $\begingroup$ You're right, it's not legal to assume all those coefficients equal $1$. But it's possible to prove that they equal $1$. Hint: If $A+B+C=0$ then $|A|=|B+C|\le|B|+|C|$. $\endgroup$ – David C. Ullrich Dec 21 '15 at 16:22
  • $\begingroup$ But here I don't have that the sum of the coefficients is $0$. How do I do that then ? $\endgroup$ – Mr. Y Dec 21 '15 at 17:03
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Consider $$(1-x)\left(x^n+\sum_{i=0}^{n-1} a_ix^i \right) = a_0 - x^{n+1} + \sum_{i=1}^n (a_i-a_{i-1})x^i\qquad (1)$$ Let $x$ with $|x|>1$ be a root of $(1).$ Therefore, $x^{n+1}=a_0 +\sum_{i=1}^{n-1} (a_i-a_{i-1})x^i $ and we have with $a_n=1,$ \begin{align} | x^{n+1}| &= \left|a_0 + \sum_{i=1}^{n} (a_i-a_{i-1})x^i\right| \\ & \le a_0 + \sum_{i=1}^n (a_i-a_{i-1})|x^i| \\ & <|x|^n\left( a_0 + \sum_{i=1}^n (a_i-a_{i-1})\right) \\ & = |x|^n\end{align} a contradiction.

Thus if there is a root such that $|x|\geq 1$ therefore $|x|=1.$

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  • $\begingroup$ Bear with me but I don't quite follow your argument.Why do I have to consider that expression in the beginning?Also what is the contradiction here ? $\endgroup$ – Mr. Y Dec 21 '15 at 16:55
  • $\begingroup$ @Mr.Y I shown $|z|$ cannot be greater than 1. $\endgroup$ – Suhail Dec 21 '15 at 16:58
  • $\begingroup$ Okay,but why do you start with that expression in the beginning $(1-x) \cdot something $ ?Where's the analogy with the expression of the problem ? $\endgroup$ – Mr. Y Dec 21 '15 at 17:06
  • $\begingroup$ What I had in mind doesn't quite work, sorry. $\endgroup$ – David C. Ullrich Dec 21 '15 at 17:11
  • $\begingroup$ @Suhail Can you elaborate a bit on the last part .I find it really hard to comprehend... $\endgroup$ – Mr. Y Dec 21 '15 at 17:38

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