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Problem:

Let $z_0$ be a root of the complex polynomial $z^n + a_{n-1}z^{n-1} + ... + a_0 $ $ (a_k \in \mathbb{C})$. Prove that $|z_0| \le \zeta$, where $\zeta$ is the only positive root of $z^n - |a_{n-1}|z^{n-1} - ... - |a_0|$.

(the preceding problem - which I've solved - was to prove that the second polynomial has in fact exactly one positive root; to be precise we'd have to assume that at least one of the $a_i$ are not equal to 0 or allow the root to be zero)

I have no idea how to approach this problem. The statement seems to be that "for given nonnegative real $a_i$ the complex polynomial with the greatest root whose coefficients are of magnitude $a_i$ is $z^n -...-a_0$", and I've tried proving this by "rotating" the coefficients one by one and observing how the roots behave, but I've had no success. (maybe it's just because I have no experience at all with complex polynomials)

I haven't studied complex analysis, so it would be great to find a solution that doesn't use results from that area. Hints would be great as well. :)

Dániel G.
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  • Suppose there is a root with modulus $|z_0| > \zeta$. What happens when you naively try to bound its modulus from below using your polynomial expression? – Travis Willse Aug 29 '14 at 06:27

2 Answers2

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If $z_0$ is a root then $-z_0^n = a_0 + \ldots + a_{n-1}z_0^{n-1}$. Now the triangle inequality says that $|z_0|^n \le |a_0| + \ldots + |a_{n-1}||z_0|^{n-1}$.

Since for positive real $x$, $x^n - |a_0| - \ldots - |a_{n-1}|x^{n-1}$ has a unique root $\zeta$, this quantity is negative for $0 \le x \le \zeta$ and positive for $x \ge \zeta$.

This shows that $|z_0| \le \zeta$.

mercio
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Here is a partial answer.

Let $f(z)=z^n - |a_{n-1}|z^{n-1} - ... - |a_0|$. when $z\to 0$, we have $f(z)<0$, when $z\to \infty$, we have $f(z)>0$. so there is a zero of $f(z)$ between 0 and $\infty$.

mike
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  • I actually did solve this problem (showing that this polynomial has exactly one positive root). Sorry for being unclear! :) – Dániel G. Aug 29 '14 at 06:35
  • Good. I forgot how to show that this is the only positive root. can you remind me? thanks! – mike Aug 29 '14 at 06:41
  • Well I've used induction on the degree of the polynomial and took the derivative. Since the derivative tends to infinity and only has one positive root, the original polynomial must be strictly increasing after it's minimal value and thus it can only have one positive root as well. (there might be other details but this was the general idea) – Dániel G. Aug 29 '14 at 06:51
  • thanks. another way to think about is that since $f(0)<$ and $f(\infty)>0$, then $f(z)$ either has single positive root or 3 positive roots, or odd number of positive roots. – mike Aug 29 '14 at 06:57