# complex polynomial

Let $$N$$ be a big integer number and consider the equation :

$$x^{N} + a_{N-1} x^{N-1} + ...+a_{1} x + o(\frac{1}{N})=0,$$ where $$o(h)$$ is by definition a term such that $$\lim_{h \to 0} o(h)/h =0$$. Assume that all coefficients $$a_j$$ have a big norm : $$|| a_j||\approx N^{j-1}$$ except $$a_{1}$$ which is asymptotically a nonzero constant.

Let $$x_s$$ be a complex root of above polynomial with the smallest norm. I want to show something similar to $$|| x_s || \approx o(\frac{1}{N})$$ or any upper bound like $$|| x_s || \leq \frac{1}{N}$$ or even smaller than that. It's obvious that if we don't have the term $$o(\frac{1}{N})$$ then smallest norm root of the polynomial $$x^{N} + a_{N-1} x^{N-1} + ...+a_{1} x=0$$ is $$x_s=0$$. So intuitively makes sense to say that for the first polynomial $$|| x_s ||$$ is also small but I can't show how small it is .

I do not expect someone gives me the exact solution of this problem because I understand we need more details, but please let me know how you tackle with these kind of questions.

• very strange definition of $O(h)$ – Fedor Petrov Dec 8 '18 at 7:32
• i've changed the $O$ into the usual $o$. – Liviu Nicolaescu Dec 8 '18 at 10:49

Denote $$x=y/N$$, your equation in terms of $$y$$ rewrites as $$N^{-1}\sum_{k=0}^{N} (a_k N^{1-k})y^{k}=0$$, you need a small root of such a polynomial in $$y$$. Denote $$a_kN^{1-k}=b_k$$, then $$b_1$$ is asymptotically constant, $$b_0$$ tends to 0 (I understand you so, please use $$o$$ instead of $$O$$ if it is the case), other $$b_i$$ are bounded (and also $$b_N$$ is very small but we do not use this.) You may use Rouchet theorem for the circle $$|y|=c$$ for small $$c$$ and the functions $$f(z)=b_1z$$, $$g(z)=\sum_{j\ne 1} b_j z^j$$. The function $$f$$ has exactly one root inside the circle $$|y|=c$$ and $$|f|>|g|$$ on the circle (for any fixed $$c>0$$ this is so if $$b_0$$ is small enough). Thus you get a root of your polynomial $$f+g$$ smaller than $$c$$.