# Inequality in Complex Plane

I'm studying numerical analysis and in the book I'm reading there is a theorem thats find a raduis such that all the roots of a polynomial $P$ (with coefficient in $\mathbb{C}$) are in the open disk with center at $(0,0)$. This can be helpful when trying to find roots as it gives a good initial guess.

The only problem I have is that I can't seem to be able to understand why this is true :

$$|P(z)| \geq |a_n z^n| - |a_{n-1}z^{n-1}+ \cdots +a_1z+a_0|.$$

($P(z)= \sum_{i = 0}^n a_i z^i$)

*also assume $|z| > 1$.

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I have edited your question to make it easier to read. Please make certain that it still reflects the question you intended to ask. –  Austin Mohr Dec 16 '11 at 7:27
Thank you very much –  Belgi Dec 16 '11 at 7:28

This is due to the reverse triangle inequality: $$|u - v| \geqslant |u| - |v|.$$ This can be derived from the usual triangle inequality as follows: $$|u| = |(u - v) + v| \leqslant |u - v| + |v|.$$

Applying the reverse triangle inequality to $u = a_nz^n$ and $v = - \left( a_{n-1} z^{n-1} + \cdots + a_1 z + a_0 \right)$ and noting that $|-v| = |v|$, we get the claim.

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Thank you for your help –  Belgi Dec 16 '11 at 7:36
A sharper version of this inequality is in my answer here –  t.b. Dec 16 '11 at 7:40
This is just the inverse triangle inequality ($|x| = |x -y + y|\leq |x-y| + |y|$, hence $|x| - |y| \leq |x-y|$).
Apply this to $x = a_n z^n$ and $y = -(a_0 + \cdots + a_{n-1} z^{n-1})$.