# proof complex inequality involves triangle inequality

Consider the polynomial $P(z) = a_0 + a_1z + a_2z ^{2} + · · · + a_nz^{n}$ where $a_j$ starts at $0$ and goes to $n$ with fixed complex numbers with $n \geq 1$ and a cannot be $0$. Show that if $P(z) = 0$ then

$$|z| ≤ \text{max} (1, {|a_0| + · · · + |a_{n−1}|\over|a_n|})$$

I'm new to complex analysis and I'm finding it hard to even interpret what the question is asking for. The only thing I can think to do is assume that $P(z)=0$ and $n$ is greater than one so I can work with the $z$'s, rather than just the constant. The inequalities kind of suggest to me that using the triangle inequality could prove useful.

## 2 Answers

First of all, this isn't really complex analysis, at least it doesn't require any tools of complex analysis.

I don't really understand what you don't understand about the question, but you may start as follows: If $P(z)=0$, then $$|z|^n=|z^{n}|=|\frac{a_0+a_1z+\dots+a_{n-1}z^{n-1}}{a_n}|$$

If $|z|\leq 1$ then the bound works. Otherwise $|z|>1$ and $P(z)=0$ implies that $$|z|=\frac{\left|-(a_0/z^{n-1} + a_1/z^{n-2} + a_2/z ^{n-3} + \cdot +a_{n-1})\right|}{|a_n|}\\\leq \frac{|a_0|/|z|^{n-1} + |a_1|/|z|^{n-2} + |a_2|/|z| ^{n-2} + \cdot +|a_{n-1}|}{|a_n|}\\\leq \frac{|a_0| + |a_1| + |a_2| + \cdot +|a_{n-1}|}{|a_n|}$$ where we used the triangle inequality and $1/|z|^k<1$. Therefore $$|z| ≤ \text{max} (1, {|a_0| + \cdots + |a_{n−1}|\over|a_n|}).$$

• In the first and second lines, should a_2/z^{n-3} follow a_1/z^{n-2}? – Kierra Aug 14 '16 at 22:44
• @Kierra Thanks! – Robert Z Aug 14 '16 at 22:57