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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.

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2  
Use your argument on h(z) = g(z+c), noting that h^(k)(0) = g^(k)(c). –  Jack Schmidt Jan 9 '11 at 5:10
    
Of course! Your hint that $h^{(k)}(0) = g^{(k)}(c)$ did it for me. I had a feeling it could be done by translation but I guess my brain was just done thinking for today. –  Antonio Vargas Jan 9 '11 at 5:41
    
@AntonioVargas How Rouches Theorem is applied here could you write an answer for this question? –  Une Femme Douce Apr 27 '13 at 11:14
1  
@Tsotsi it has been a while but I will see if I can figure it out again :) –  Antonio Vargas Apr 27 '13 at 15:01
    
@Tsotsi, I've just posted my answer. –  Antonio Vargas Apr 27 '13 at 16:41

1 Answer 1

up vote 3 down vote accepted

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 $|z-c| < \delta$.

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Thank you very much –  Une Femme Douce Apr 27 '13 at 21:57

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