# How to prove this inequality of polynomial in complex number

the inequaltiy in this picture

I try to prove it but i stuck with $\frac2{|a_n|}$. How did he got it ??

I followed the proof in the textbook but the term $\frac{2}{|a_n|}$ does not make sense to me.

It said $$|w| \leq \sum_{i=0}^{n-1} \frac{|a_i|}{|z^{n-i}|}.$$ As $R$ increases, we have $$|w| < n\frac{|a_n|}{2n}$$ and this step which i did not get it

• Which inequality? Also, consider following a MathJax tutorial to format your post. – Glorfindel May 3 '17 at 16:38
• i edited the description and i included the the inequality in the link imgur.com/a/4TJ9O – NSG May 3 '17 at 16:48

As $|z| \rightarrow \infty$, $\frac{|a_i|}{|z|^{n-i}} \rightarrow 0$ for $0\leq i<n$

That is when $|z|$ is sufficiently large, we can make $\frac{|a_i|}{|z|^{n-i}}$ arbitrarily small in magnitude.

That is $\forall \epsilon > 0$, $\exists R_{\epsilon, i} > 0$, $|z| > R_{\epsilon,i} \implies \frac{|a_i|}{|z|^{n-i}}<\epsilon.$

Since $a_n \neq 0$, we can let $\epsilon = \frac{|a_n|}{2n}>0$ and we know that

$\exists R_{ i} > 0$, $|z| > R_{i} \implies \frac{|a_i|}{|z|^{n-i}}<\frac{|a_n|}{2n}.$

Pick the same $R$ for all the $i \in \{ 0, \ldots , n-1 \}$, for example, let $R= \max \{ R_0, \ldots, R_{n-1}\}$.

hence if $|z|>R$, $$|w|\leq \sum_{i=0}^{n-1}\frac{|a_i|}{|z|^{n-i}} < \sum_{i=0}^{n-1}\frac{|a_n|}{2n} = n \frac{|a_n|}{2n}=\frac{|a_n|}{2}$$

• still i didn't how we reach to 2nd step abs(ai)\ 2n – NSG May 3 '17 at 17:19
• Let's forget about complex number for now, and go back to real analysis. Do you agree that $\forall \epsilon > 0$, I can pick $R$ large enough such that $\frac1x < \epsilon$, if $x > R$? – Siong Thye Goh May 3 '17 at 17:23
• yes i agree with you in this – NSG May 3 '17 at 17:31
• awesome, the same thing should also work on $\frac{3}{x^2}, \frac{5}{x^3}, \frac{0}{x^7},\ldots, \frac{\pi}{x^n}$, our $\epsilon$ is equal to $\frac{|a_n|}{2n}$. – Siong Thye Goh May 3 '17 at 17:35
• remark: those function are examples of nonnegative functions which converges to $0$. – Siong Thye Goh May 3 '17 at 17:43