proof $x^m-10x^{m-1} + 1 = 0 \implies |x| < 10$ I'm trying to prove that if $x^m-10x^{m-1} + 1 = 0$, then $|x| < 10$. This is obvious if x is real, but I don't know how to prove it if x can be complex. I tried to implement $x = r(\cos(\theta) + i\sin(\theta))$, with $r \ge 10$, but I couldn't find a contradiction. Any help would be very much appriciated.
 A: If $m=1$, it's clear. If $m=2$, it is a quadratic equation with discriminant $\Delta = 96$ so the roots are $5 \pm 2\sqrt{6} \in [-10,10]$. Assume $m \geq 3$.
Write $x=|x|e^{i\theta}$ a root with modulus at least $10$ and $|\theta| \leq \pi$, so that $x=10-x^{1-m}$. Write $x^{1-m}=u+iv$, with $u^2+v^2 = |x|^{2-2m} \leq 10^{-4}$.
Thus an argument of $x$ is $-\tan^{-1}\frac{v}{10-u}$ ie $0 \leq |\theta| \leq \frac{|v|}{10-u} \leq \frac{10^{1-m}}{9.99}$. In particular an argument $\alpha$ of $x^{1-m}=u+iv$ is such that $|\alpha| \leq \frac{(m-1)10^{1-m}}{9.99}$, so that for $m \geq 3$,  $|\alpha| \leq \pi/10$, and thus $|x|^2 = 100-2u+u^2+v^2 = 100+|x|^{2-2m}-2|x|^{1-m}\cos{\alpha} \leq 100+|x|^{1-m}(0.01-2\cos{\alpha}) \leq 100+|x|^{1-m}(0.01-2\cos(\pi/10)) < 100$, so we get a contradiction.
A: This may be killing a fly with a sledgehammer given the tags, but we can employ Rouché's theorem with $f(x) = x^m - 10x^{m-1} + 1$ and $g(x) = -10x^{m-1}$ on the circle $|x| = R$ ($x \in \mathbb C$), where $R$ is some real constant that we will pick.
Here is the theorem:
Theorem (Rouché). Let $U$ be an open subset of $\mathbb C$ and let $\gamma$ be a closed path homologous to 0 in $U$. Let $f$ and $g$ be analytic complex functions on $U$ such that $|f(z) - g(z)| < |g(z)|$ for all $z$ on $\gamma$. Then $f$ and $g$ have the same number of zeros in the interior of $\gamma$.
This definition is pretty dense, so let's state the specific case of the theorem that we need here:
Theorem (Rouché, Free Trial Version). Let $f$ and $g$ be polynomials. Suppose that for all $x \in \mathbb C$ such that $|x| = R$, we have $|f(x) - g(x)| < |g(x)|$. Then $f$ and $g$ have the same number of roots (counting multiplicity) with magnitude less than $R$.
The primary hypothesis of Rouché's theorem is that $|f(x) - g(x)| < |g(x)|$ when $|x| = R$. Note that when $|x|=R$, we have
$$|g(x)| = |-10x^{m-1}| = 10R^{m-1} \hspace{2cm} |f(x) - g(x)| = |x^m+1| \leq R^m+1$$
This means that the hypothesis of the theorem holds as long as $R^m+1 \leq 10R^{m-1}$, or rearranging, as long as
$$f(R) = R^m-10R^{m-1}+1 < 0$$
Now, you've solved the problem in the real case, so you have likely been able to show that your polynomial $f(x)$ has a root $R'$ of multiplicity 1 between 0 and 10. Otherwise, observe that the only multiple root of the derivative of $f$ is 0 (there are other ways to show this without calculus; I'll leave it as an exercise). Since $f(x) \to \infty$ as $x \to \infty$, we know that for $R$ "just less" than $R'$, we have $f(R) < 0$, so we can apply Rouché's theorem.
Rouché gives us that $f(x)$ and $g(x)$ have the same number of roots with magnitude less than $R$. Since $g(x) = -10x^{m-1}$ has $m-1$ roots in this ball (in the form of the single root 0, which has multiplicity $m-1$), so does $f(x)$. Now, since $f(x)$ is a polynomial of degree $m$, we know the last root of $f$ is $R'$. This means that $f(x)$ has $m-1$ roots of magnitude less than $R$, and one root of magnitude $R'$. Since $R < R' < 10$, this completes the proof.
