How is the triangle inequality used to derive this inequality?

I am trying to prove that any non-constant complex polynomial tends to infinity as z goes to infinity. Someone asked this question on this website here. In that question, the following hint is given: $\def\abs#1{\left|#1\right|}$ $$\abs{\sum_{k=0}^n a_k z^k} \ge \abs{a_n}\abs{z}^n - \sum_{k=0}^{n-1}\abs{a_k}\abs z^k$$

They say that this is a consequence of the triangle inequality, but I am not seeing how. Any help would be great, thanks.

• that really has absolutely nothing to do with complex analysis i am afraid – Dominic Michaelis Mar 19 '13 at 23:24

$$\def\abs#1{\left|#1\right|} |a_nz^n|= \abs{\sum_{k=0}^n a_k z^k - \sum_{k=0}^{n-1}a_kz^k}\leq \abs{\sum_{k=0}^n a_k z^k}+\abs{\sum_{k=0}^{n-1}{a_k}z^k} \leq \abs{\sum_{k=0}^n a_k z^k}+\sum_{k=0}^{n-1}\abs{a_k}\abs{z^k}$$
$$\def\abs#1{\left|#1\right|}|a_nz^n|= \abs{\sum_{k=0}^n a_k z^k - \sum_{k=0}^{n-1}a_kz^k}\leq \abs{\sum_{k=0}^n a_k z^k}+\sum_{k=0}^{n-1}\abs{a_k}\abs z^k$$
• Small typo --- $z_n$ meant to be $z^n$. – Gerry Myerson Mar 19 '13 at 23:26
The general idea here is to "reverse" the triangle inequality by adding and subtracting terms. In particular, $$|x| = |x + y - y| \le |x + y| + |y|$$ which implies $$|x| - |y| \le |x + y|$$ See if you can insert appropriate quantities for $x,y$ to apply this "reverse" triangle inequality to your problem.