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Let $p(x)$ be a polynomial with complex coefficients and $p(\tilde x)=0$. Choose $\delta>0$ small enough, such that $\tilde x$ is the only root of $p$ in $B_\delta(\tilde x)$. I want to show that there exists a constant $c$ such that for all $x\in\partial B_{\frac{\delta}{\lambda}}(\tilde x)$ with $\lambda\geq 1$ there holds

$$\vert p(x)\vert\geq c\lambda^{-1}$$.

I'm thankful for every kind of help :)

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  • $\begingroup$ Presumably you mean $|p(x)| \ge c \lambda^{-1}$. $\endgroup$ – Robert Israel Apr 22 '16 at 18:44
  • $\begingroup$ yes,! sorry for that $\endgroup$ – lbf_1994 Apr 22 '16 at 18:45
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It's not true. Try $p(x) = x^2$. Or do you mean that $\overline{x}$ is a simple root?

EDIT: With the assumption that $\overline{x}$ is a simple root, $$0 \ne p'(\overline{x}) = \lim_{x \to \overline{x}} \dfrac{p(x)}{x-\overline{x}}$$ Take $c = |p'(\overline{x})|/2$. There is some $\eta > 0$ such that for $|x - \overline{x}| \le \eta$ we have $$\left| \dfrac{p(x)}{x-\overline{x}} - p'(\overline{x}) \right| < c$$ and thus $|p(x)| \ge c |x - \overline{x}|$. You want to take $1/\lambda \le \eta$.
(I presume you meant "... for some $\lambda$ ...". It certainly can't work for all $\lambda \ge 1$: e.g. you might have another zero at distance $1$ from $\overline{x}$.

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  • $\begingroup$ I guess 'the only root' indicates it should be simple... $\endgroup$ – Thomas Apr 22 '16 at 18:47
  • $\begingroup$ yes it should be a simple root. $\endgroup$ – lbf_1994 Apr 22 '16 at 18:50

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