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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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1 Answer 1

up vote 4 down vote accepted

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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Got it! Thanks for the hint: $|P(z)| = |(z^2 + a_1z) + a_0| \geq |z^2 + a_1z| - |a_0| \geq |z|^2 - |a_1||z| - |a_0|$ –  ItsNotObvious May 17 '11 at 21:01
    
OR (same proof actually), you can use the triangle inequality for $z^2= P(x)-a_1z-a_0$. –  N. S. May 18 '11 at 1:29

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