2
$\begingroup$

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

$\endgroup$
4
  • $\begingroup$ 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
  • 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
  • $\begingroup$ 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
  • $\begingroup$ +1 Good question (and my apology) $\endgroup$ – Dr. Wolfgang Hintze Jan 17 at 10:38
4
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ Whoops sorry I was typing and didn't notice your answer. I see that our answers are basically the same. Do you mind if I keep my answer there? $\endgroup$ – DuduBob Jan 17 at 10:39
  • $\begingroup$ @DuduBob: Sure, no worries. It confirms the approach. $\endgroup$ – quasi Jan 17 at 10:41
  • $\begingroup$ Thank you for your solution $\endgroup$ – Ars Jan 17 at 10:42
1
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ Thank you for your solution $\endgroup$ – Ars Jan 17 at 10:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.