Trouble with a problem involving Rouché's Theorem The problem is from Marden, the first section:
The polynomial $g(z) = z^n + b_1 z^{n-1} + ... + b_n$ has at least $m+1$ zeros in an arbitrary neighborhood of a point $z = c$ if $|g^{(k)}(c)| \leq \epsilon$ for $k = 0,1,...,m$ and for $\epsilon$ sufficiently small and positive.
There is a hint provided: use Rouché's Theorem.
I can prove the result in the special case of $c = 0$, because then I can bound each of the relevant $b_j$ by $\epsilon$. Unfortunately I don't see a way to extend this to the general case.
I would appreciate some help on working toward a solution. I've been stumped since lunch on this one.
 A: Since $g(z)$ is a polynomial of degree $n$, so is its Taylor series about the point $z=c$.  Note that $g^{(n)}(z) = n!$, so that the coefficient of $(z-c)^n$ in this series is $1$:
$$
g(z) = g(c) + g'(c)(z-c) + \cdots + \frac{g^{(n-1)}(c)}{(n-1)!} (z-c)^{n-1} + (z-c)^n.
$$
Suppose that, for some $m \in \{0,1,2,\cdots,n-1\}$ and $\epsilon > 0$, we know that
$$
\left|g^{(k)}(c)\right| \leq \epsilon
$$
for all $0 \leq k \leq m$.  It then follows from the triangle inequality that the head of the polynomial satisfies
$$
\begin{align}
\left|g(c) + g'(c)(z-c) + \cdots + \frac{g^{(m)}(c)}{m!} (z-c)^{m}\right| &\leq \epsilon \sum_{k=0}^{m} \frac{|z-c|^k}{k!} \\
&< \epsilon e^{|z-c|}.
\tag{1}
\end{align}
$$
The rest of the polynomial
$$
h(z) = \frac{g^{(m+1)}(c)}{(m+1)!} (z-c)^{m+1} + \cdots + \frac{g^{(n-1)}(c)}{(n-1)!} (z-c)^{n-1} + (z-c)^n
$$
has a zero of multiplicity at least $m+1$ at $z=c$, and, since the zeros of analytic functions are isolated, this is the only zero of $h(z)$ in the closed disk $|z-c| \leq \delta$ for all $\delta > 0$ small enough.  The circle $|z-c| = \delta$ is compact and $h(z) \neq 0$ there, so we can also find a $\lambda > 0$ such that
$$
|h(z)| \geq \lambda > 0 \qquad \text{on}\,\,\, |z-c| = \delta.
\tag{2}
$$
Now, if we choose
$$
\epsilon \leq \lambda e^{-\delta}
$$
then we can deduce from $(1)$ and $(2)$ that
$$
\begin{align}
\left|g(c) + g'(c)(z-c) + \cdots + \frac{g^{(m)}(c)}{m!} (z-c)^{m}\right| &< \epsilon e^{\delta} \\
&\leq \lambda \\
&\leq |h(z)|
\end{align}
$$
on the circle $|z-c| = \delta$.
We may now apply Rouché's to conclude that for any fixed $\delta > 0$ we can find an $\epsilon > 0$ small enough so that the polynomial $g(z)$ has at least $m+1$ zeros in the disk $|z-c| < \delta$.
