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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. $|q(z)| \geq R^2 - |a|R - |b| > R^2/2.$

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    Sorry about that. How embarra$s$sing. I've edited the answer to reflect your comments.2011-12-02