# Roots of cubic polynomial lying inside the circle

Show that all roots of $a+bz+cz^2+z^3=0$ lie inside the circle $|z|=max{\{1,|a|+|b|+|c| \}}$

Now this problem is given in Beardon's Algebra and Geometry third chapter on complex numbers.

What might be relevant for this problem:

• author previously discussed roots of unity;
• a little (I mean abut a page of informal discussion) about cubic and quartic equations;
• then gave proof of fundamental theorem of algebra (the existence of root was given as a informal proof and rest using induction) and then the corollary of it (If $p(z) = q(z)$ at $n + 1$ distinct points then $p(z) = q(z)$ for all $z$, where both polynomials are of degree at most $n$);

I was trying to see how I should approach it with no success for quite some time. Looked on the net and found, that this would be kind of easy with Rouche's theorem, but I was not given that. So is it possible to solve it in a simple way with what was given? Thanks!

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## 2 Answers

Suppose $a + bz + cz^2 + z^3 = 0$ and $|z| > \max\{1,|a|+|b|+|c|\}$. Use the triangle inequality to show that

$$|z^3| = |a + bz + cz^2| < (|a|+|b|+|c|)|z^2|,$$

which contradicts the assumption on the size of $|z|$.

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I'd simply be looking at showing that the $z^3$ term was dominant, so there could be no roots beyond the bound. I don't think it is at all sophisticated.

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