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Given the complex polynomial $P(z) = z^2 + a_1z + a_0$ and the constraint that $|z| > 1$, I'm trying to show that $|P(z)| \geq |z|^2 - |a_1||z| - |a_0|$. The obvious thing to do here of course is to apply the triangle inequality which yields $ |P(z)| = |z^2 + a_1z + a_0| \leq |z|^2 + |a_1||z| + |a_0| $ Just as obvious though is the fact that this is not remotely close to what I am trying to show. I'm sure I'm missing a relevant detail. I have attempted, and failed, to productively apply the constraint on $|z|$. This is effectively the start of a proof of the Fundamental Theorem of Algebra.

Would appreciate any tips.

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Hint. The triangle inequality also shows that $|a+b|\geq |a|-|b|.$

To see this, note that $|a| = |a+b-b| \leq |a+b|+|b|.$

(In point of fact, I don't see that the constraint is needed at all to establish what you want, though it would be useful to obtain further lower bounds on $|P(z)|$...)

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    OR (same proof actually), you can use the triangle inequality for $z^2= P(x)-a_1z-a_0$.2011-05-18