# Inequality question (with complex numbers)

I was wondering how would you derive/get $|q(z)|\ge R^2-|a|R-|b|>R^2/2$?

Thanks.

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By application of the triangle inequality, we have $$|q(z)| = |z^2+az+b| \geq |z^2| - |az| - |b| = |z|^2 - |a||z| - |b|.$$ Edited: fixed triangle inequality mistake
Now if we take $R \geq |z|$ and $R$ big enough so that $R^2 \geq |a|R + |z|^2-|a||z|$, we have $$|q(z)| \geq |z|^2 - |a||z| - |b| \geq R^2 - |a|R - |b|.$$ Then since $R^2/2$ is eventually bigger than $|a|R + |b|$ for large $R$, we can add the negative number $|a|R + |b| - R^2/2$ to the right hand side to get a strict inequality $$R^2 - |a|R - |b| > R^2 - |a|R - |b| + (|a|R + |b| - R^2/2) = R^2/2.$$ Putting these two pieces is your inequality. $$|q(z)| \geq R^2 - |a|R - |b| > R^2/2.$$
 Thanks, can I ask why $|z^2+az+b| \geq |z^2| + |az| + |b|$? Because isnt the triangle inequality $|x+y|<|x|+|y|$? – Ray Dec 2 '11 at 2:46 Dane, you have the first inequality the wrong way round. Ray, a variant of the triangle inequality says $|a+b|\ge|a|-|b|$. – Gerry Myerson Dec 2 '11 at 2:55 The variant is the stronger $|a+b| \ge ||a|-|b||$. Since $|x| > x$ for all real $x$, we also have $||a|-|b|| \ge |a|-|b|$ – Aleks Vlasev Dec 2 '11 at 3:30 Sorry about that. How embarrassing. I've edited the answer to reflect your comments. – Dane Dec 2 '11 at 17:18