If $m$ is a zero of $$ p(x)=x^n+a_{n-1} x^{n-1}+\ldots+a_0, $$ where $a_i$ may be complex, show that $$ |m| \leq \max \left\{1,\left|a_0\right|+\left|a_1\right|+\ldots+\left|a_{n-1}\right|\right\} $$ I am not sure where to begin on this problem. I think there is something to with symmetric functions, but I don't know what to do. Any help will be appreciated. Thank you.
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HINT. Take a look at $\frac{dp}{dx}$. – Paul Tanenbaum Nov 10 '23 at 15:45
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I am still confused, but do we proceed by plugging in m in the derivative and taking absolute value on both sides? – Anomaly Nov 10 '23 at 15:55
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Suppose that $m$ is a root with $|m| > 1$. Then we need to show that $|m| \le |a_0| + \dots + |a_{n-1}|$. But note that, since $m^n = -\Big( \sum_{i=0}^{n-1} a_i m^i\Big)$,
$|m|^n = \Big| \sum_{i=0}^{n-1} a_i m^i \Big|$
Now try bounding the right hand side using the triangle inequality.
In fact this is
$|m|^n \le \sum_{i=0}^{n-1} |a_i| |m|^{i} \le |m|^{n-1} \sum_i |a_i|$
from which is quickly follows.
Sam Moore
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