Let \(a,b,c, z\) be complex numbers such that \(|a| = |b| = |c| > 0 \) and \(az^2+bz+c=0\)

Find the largest possible value of \(|z|\).

AoPS.Morrisville Nov 18, 2019

#1**0 **

Let a = b = c. Then the quadratic is az^2 + az + a = a(z^2 + z + 1) = 0. The roots of z are -1/2 \pm sqrt(3)/2*i, and these have magnitude 1, so this is the maximum.

Guest Nov 18, 2019

#2**0 **

I don't believe we can assume that a = b = c, since all we are given is that their three moduli are equal and nonzero. So I will provide an answer that does not reqire that assumption.

We are given that \(az^2+bz + c = 0\), so we can assume

\(0= |az^2+bz+c| \geq|az^2+bz|-|c| \geq|az^2|-|bz|-|c|\).

But then

\(|a||z|^2-|b||z|-|c|=k|z|^2-k|z|-k=k(|z|^2-|z|-1)\leq0\),

where \(k=|a|=|b|=|c|\)is nonzero and positive.

This forces the inequality \(|z|^2-|z|-1\leq0\).

From the graph of the real-valued function \(f(|z|)=|z|^2-|z|-1\) given above, one can see that the values of \(|z|\) for the portion of the graph __below and touching the \(|z|\) axis__ range from

|z|= \(\frac{1-\sqrt{5}}{2} \) to |z|=\(\frac{1+\sqrt{5}}{2} \). So the largest value of |z| is \(\frac{1+\sqrt{5}}{2} \).

Gadfly Nov 18, 2019