# How to show that it holds $|z|<2\max_{0\le k<n}|a_k|^{\frac{1}{n-k}}$ for any root of $X^n+\sum_{k=0}^{n-1}a_kX^k$?

Let $z\in\mathbb{C}$ be a root of the complex polynomial $$f=X^n+\sum_{k=0}^{n-1}a_kX^k$$ I want to show that it holds $$|z|<2\max_{0\le k<n}|a_k|^{\frac{1}{n-k}}$$ Proof: For $s>1$, consider $$f_s:=\frac{1}{s^n}f(sX)=X^n+\sum_{k=0}^{n-1}\frac{a_k}{s^{n-k}}X^k$$ Evaluation of $f_s$ at $z$ yields \begin{equation} \begin{split} f_s(z) &= \sum_{k=0}^{n-1}\frac{a_k}{s^{n-k}}z^k-\sum_{k=0}^{n-1}a_kz^k \\ &=\sum_{k=0}^{n-1}\left(\frac{1}{s^{n-k}}-1\right)a_kz^k \end{split} \end{equation} Since $\sqrt[n-k]{2}\in (1,2]$ and $s>1$ it holds $$\frac{1}{\sqrt[n-k]{2}}<s\;\;\;\Leftrightarrow\;\;\;\frac{1}{2}<s^{n-k}\;\;\;\Leftrightarrow\;\;\;\frac{1}{s^{n-k}}-1<1$$ Thus, \begin{equation} \begin{split} |f_s(z)| &\le \sum_{k=0}^{n-1}\left|\frac{1}{s^{n-k}}-1\right||a_k||z|^k \\ &\le \sum_{k=0}^{n-1}|a_k||z|^k \\ &< \sum_{k=0}^{n-1}|a_k|M^k \end{split} \end{equation} where $$M:=1+\max_{0\le k<n}|a_k|$$ is an upper bound for the length of any root of $f$ by Cauchy's theorem. Now there comes the point where I'm unsure how I should proceed. I would be really thankful if someone could help me out.

By the triangle inequality we have

$$|f(z)| \geq |z|^n - \sum_{k=0}^{n-1} |a_k| |z|^k,$$

so if $z$ is a root of $f$ then $|z| \leq r$, where $r$ is the unique positive root of the equation

$$x^n - \sum_{k=0}^{n-1} |a_k| x^k = 0.$$

(Note that Descartes' rule of signs implies that the equation does indeed have a unique positive root.) Let

$$M := 2 \max_{0 \le k < n} |a_k|^{1/(n-k)}.$$

For all $k$ we have

$$2^{n-k} |a_k| \leq M^{n-k},$$

so

$$\sum_{k=0}^{n-1} |a_k| M^k \leq M^n \sum_{k=0}^{n-1} 2^{k-n} = M^n (1-2^{-n}) < M^n.$$

Thus $M > r$, from which the result follows.

• The essence of this argument is drawn from the proof of Theorem 30,2 of Marden's Geometry of Polynomials. – Antonio Vargas May 12 '14 at 9:03
• Thanks for your detailed answer. But shouldn't it be $$M^n(1-2^{-n})$$ in the last equation? In addition, I think the triangle inequality yields $$|f|\ge \left||X|^n-\sum_{k=0}^{n-1}|a_k||X|^k\right\|$$ However, I'm unsure if you argue with the knowledge that $z$ is a root of $f$ (cause $z$ was introduced in the question as a root of $f$) or you placed $z$ as an arbitrary variable. – 0xbadf00d May 12 '14 at 16:31
• Whoops, you're right, it should be $2^{-n}$, thanks for spotting that. And $$\left||z|^n - \sum_{k=0}^{n-1}|a_k||z|^k\right| \geq |z|^n - \sum_{k=0}^{n-1}|a_k||z|^k,$$ no? – Antonio Vargas May 12 '14 at 16:32
• You're right, you've used the triangle inequality in the right way. My mistake. – 0xbadf00d May 12 '14 at 16:36