Roots of polynomials with a specific coefficient 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
 A: 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.
A: 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.
