Please prove in the following polynomial $$z^m-az^{m-1}-b=0 \hspace{1cm} z \in \mathbb{C} \hspace{1cm} m \in \mathbb{N}$$ if $$|a|>2$$ then holds at least one of the roots in the$$|z|>1$$ I tried to use the following relation but I did not succeed. $$\sum_{ i=1}^m {z_i} = -\frac{-a}{1}=a$$ If you have an idea to prove it, please say. thank you
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$\begingroup$ The assertion is not true in general. Example: with $m=2$, $b=-(\frac{a}{2})^2$ the two solutions coincide and are given by $x_s = \frac {a}{2}$. Obviously $|x_s|$ can be arbitarily large. $\endgroup$ – Dr. Wolfgang Hintze Jan 17 at 10:01
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1$\begingroup$ Dear Hintze, What you said also confirms what I said. $$x_s=\frac{a}{2}$$ so if $$|a|>2$$ Then there is at least one root that $$|x_s|>1$$ $\endgroup$ – Ars Jan 17 at 10:12
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$\begingroup$ You are right, and I was wrong, sorry. Obviously too early in the morning for me :-( $\endgroup$ – Dr. Wolfgang Hintze Jan 17 at 10:20
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$\begingroup$ +1 Good question (and my apology) $\endgroup$ – Dr. Wolfgang Hintze Jan 17 at 10:38
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Let $m > 1$ be an integer, let $a,b\in\mathbb{C}$ be such that $|a| > 2$, and let $z_1,...,z_m\in \mathbb{C}$ be the roots (repetitions allowed) of the equation $$z^m-az^{m-1}-b=0$$ We want to show $|z| > 1$ for some $z\in\{z_1,...,z_m\}$.
If $b=0$, then $z=a$ is a root, and we're done.
So assume $b\ne 0$.
Then $z_1,...,z_m$ are all nonzero.
Suppose $|z| \le 1$ for all $z\in\{z_1,...,z_m\}$. \begin{align*} \text{Then}\;\;& z\in\{z_1,...,z_m\} \\[4pt] \implies\;& z^m-az^{m-1}-b=0 \\[4pt] \implies\;& z^{m-1}(z-a)=b \\[4pt] \implies\;& |z|^{m-1}|a-z|=|b| \\[4pt] \implies\;& |z|^{m-1}(|a|-|z|)\le|b| \\[4pt] \implies\;& |z|^{m-1}(2-1) < |b| \\[4pt] \implies\;& |z|^{m-1}|z| < |b| \\[4pt] \implies\;& |z|^m < |b| \\[4pt] \implies\;& |z| < |b|^{\large{{\frac{1}{m}}}} \\[4pt] \implies\;& \prod_{i=1}^m |z_i| < |b| \\[4pt] \implies\;& \left|\prod_{i=1}^m z_i\right| < |b| \end{align*} contradiction.
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This is an application of the triangle inequality. You are working in $\mathbb{C}$, so we have $$z^m-az^{m-a}-b = (z-\alpha_1) \cdots (z-\alpha_m) = 0$$ And $\alpha_1, \cdots, \alpha_m$ are the roots. Now suppose $|\alpha_i| < 1$ for all $i$. Note that $$b = \alpha_1 \cdots \alpha_m.$$ But we have $$z^m-az^{m-1} = b$$ which implies $$|a||z^{m-1}| - |z^m| \leq |z^m-az^{m-1}| = |b|$$ i.e. $$(2-|z|)|z^{m-1}| \leq |b|$$ Now take $z = \alpha_0$ to be the root with the largest modulus. Then $$(2-|\alpha_0|)|\alpha_0|^{m-1} \leq |\alpha_1| \cdots |\alpha_m|$$ Now every term on the left-hand side is greater than or equal to every term on the right, which $(2-|\alpha_0|) > 1$ strictly greater than any term on the right. This is a contradiction.
