Using Rouché's theorem to find a compact domain for the roots of a polynomial Im trying to solve a problem X.12.6 from Sarason's Complex Function Theory:

So I tried using Rouché's theorm, on the function $g(z)=z^n$, on the domain $\{z:z<\sqrt{1+|c_{n-1}|^2+...+|c_0|^2}\}$ and using the fact that in that open circle $g$ has $n$ roots ($0$ is zero of order $n$) and then get by the thorem that the polynomial $f$ has n roots in that circle.
The problem is that I cant show that $|f(z)-g(z)|<|g(z)|$ on the boundary of the circle.
Can you please give me a hint how to show that?
 A: Let $z$ be a root of the polynomial. If $|z| \le 1$, then the root clearly lies in the circle with radius $ \sqrt{1 + |c_0|^2 + \dots + |c_{n-1}|^2}$, so the statement is (trivially) true.
So let's take a root $z$ such that $|z| > 1$ (if no such root exists, then the statement is proved)
Let $z^n + c_{n-1}z^{n-1} + \dots + c_1z + c_0 = 0$.  Then you get $z^n = - (c_{n-1}z^{n-1} + \dots + c_1z + c_0)$. Taking the modulus of both sides $\left|z\right |^n = \left| c_{n-1}z^{n-1} + \dots + c_1z + c_0 \right|$.
Square both sides to get $\left|z\right |^{2n} = \left|c_{n-1}z^{n-1} + \dots + c_1z + c_0\right|^2$.
Now use the Cauchy Schwarz inequality to get $$\left|z\right|^{2n} = \left|c_{n-1}z^{n-1} + \dots + c_1z + c_0\right|^2 \le \left(1 + \left|z\right|^2 + \dots + \left|z\right| ^ {2n-2}\right) \left(\left|c_0\right|^2 + \dots + \left|c_{n-1}\right|^2\right) = $$
$$= \frac{\left|z\right|^{2n} - 1}{\left|z\right|^2 - 1}\left(\left|c_0\right|^2 + \dots + \left|c_{n-1}\right|^2\right)$$
A little bit of algebra yields $$\left|z\right|^2-1 \le \frac{\left|z\right|^{2n}-1}{\left|z\right|^{2n}}\left(\left|c_0\right|^2 + \dots + \left|c_{n-1}\right|^2\right)$$
And since $\displaystyle \frac{|z|^{2n}-1}{|z|^{2n}}\le 1$, you finally get $$|z|^2 \le 1 + |c_0|^2 + \dots + |c_{n-1}|^2$$
that is, if $z$ is a root of a the polynomial $P(z) = z^n + c_{n-1}z^{n-1}+\dots + c_1z + c_0$, then
$$|z| \le \sqrt{1 + |c_0|^2 + \dots + |c_{n-1}|^2}$$
A: Hint: let $p = \sqrt{|c_{n-1}|^2 + \ldots + |c_0|^2}$ and $|z| = r$.  If
$r^2> 1 + p^2$ then
$$r^{2n} >  \dfrac{p^2 r^{2n}}{r^2-1} > p^2 (1 + r^2 + \ldots + r^{2n-2}) $$
Now use Cauchy-Schwarz to estimate $|c_0 + c_1 z + \ldots + c_{n-1} z^{n-1}|$ when $|z| = r$.
