Show that all the solutions to $z^n + a_{n-1}z^{n-1}+ \dots + a_1z + a_0 = 0$ satisfy $|z| < c$. 
Let $a_0, \dots, a_{n-1} \in \Bbb C$. Show that if $c \in (0, \infty)$ is such that $$|\frac{a_{n-1}}{c}| + \dots + |\frac{a_{1}}{c^{n-1}}| + |\frac{a_{0}}{c^n}| < 1,$$ then all the solutions to $z^n + a_{n-1}z^{n-1}+ \dots + a_1z + a_0 = 0$ satisfy $|z| < c$.

Is there a geometric intuition for this problem? I think it's trying to say that if I can find a positive real number $c$ such that the sum of the scaled coefficients (except the first $1$) of $z^n + a_{n-1}z^{n-1}+ \dots + a_1z + a_0$ is less than $1$, then it must be that the solutions to this equation have modulus less than $c$ which would mean that they lie inside the open disc with radius $c$.
I think I should think of this $z^n + a_{n-1}z^{n-1}+ \dots + a_1z + a_0$ as a map $\Bbb C \to \Bbb C$ and try to figure out why the inputs that map to $0$ from the domain should lie on the open disc with radius $c$, but I don't know why this should be true. This was a question from an algebraic topology book so I suspect it should have something to do with the way we prove the fundamental theorem of algebra using fundamental groups.
 A: If $|\frac{a_{n-1}}{c}| + \dots + |\frac{a_{1}}{c^{n-1}}| + |\frac{a_{0}}{c^n}| < 1$, then for $z$ such that $|z| \ge c$,
$|z^n| > |\frac{a_{n-1}}{c}||z^n| + \dots + |\frac{a_{1}}{c^{n-1}}||z^n| + |\frac{a_{0}}{c^n}||z^n|$
$|z^n| > |a_{n-1}||z^{n-1}| + \dots + |a_{1}||z| + |a_{0}| \ge |a_{n-1}z^{n-1} + \dots + a_{1}z + a_{0}|$
so $z^n + a_{n-1}z^{n-1}+ \dots + a_1z + a_0 = 0$ is not possible for $|z| \ge c$.
A: If
$ z^n + a_{n-1}z^{n-1} + ... + a_1z + a_0 = 0$ we have
$$ |z^n| = |-a_{n-1}z^{n-1} - ... - a_1z - a_0| \leq |z^n|(|\frac{a_{n-1}}{z}| + ... + |\frac{a_1}{z^{n-1}}| + |\frac{a_0}{z^n}|)$$
then
$$ |\frac{a_{n-1}}{c}| + ... + |\frac{a_1}{c^{n-1}}| + |\frac{a_0}{c^n}| < 1 \leq |\frac{a_{n-1}}{z}| + ... + |\frac{a_1}{z^{n-1}}| + |\frac{a_0}{z^n}|$$
hence it follows that $|z| < c$.
A: There is a well-known root radius estimate that says that if
$$
R=\max(1,|a_0|+|a_1|+...+|a_{n-1}|),
$$
then all roots satisfy $|z|<R$.
If the coefficient are not of an equal small size, this might not be the best estimate. So one idea is to apply this same estimate to the polynomial $q(w)=p(cw)$. To get again a normalized polynomial, one has to divide by the leading coefficient, so the roots of $q$ satisfy
$$
0=w^n+\frac{a_{n-1}}{c}w^{n-1}+\frac{a_{n-2}}{c^2}w^{n-2}+...+\frac{a_0}{c^n}
$$
with root radius estimate
$$
|w|\le R=\max\left(1,\frac{|a_{n-1}|}{c}+\frac{|a_{n-2}|}{c^2}+...+\frac{|a_0|}{c^n}\right)
$$
Now if the second expression in the maximum is smaller than the first, then $R=1$ and all roots of $p$ satisfy $|z|=c|w|\le c$.
