HMMT, 2014

Algebra Difficulty 5.2 AIME, harder Prove it United States

Problem:
Given that $w$ and $z$ are complex numbers such that $|w+z|=1$ and $|w^{2}+z^{2}|=14$ , find the smallest possible value of $|w^{3}+z^{3}|$ . Here, $|\cdot|$ denotes the absolute value of a complex number, given by $|a+b i|=\sqrt{a^{2}+b^{2}}$ whenever $a$ and $b$ are real numbers.

Solution

Solution:
$|w^{3}+z^{3}| = |w+z|\,|w^{2}-wz+z^{2}| = |w^{2}-wz+z^{2}| = \left|\frac{3}{2}(w^{2}+z^{2})-\frac{1}{2}(w+z)^{2}\right|$ .

By the triangle inequality,
$\left|\frac{3}{2}(w^{2}+z^{2})-\frac{1}{2}(w+z)^{2}+\frac{1}{2}(w+z)^{2}\right| \leq \left|\frac{3}{2}(w^{2}+z^{2})-\frac{1}{2}(w+z)^{2}\right| + \left|\frac{1}{2}(w+z)^{2}\right|.$
By rearranging and simplifying, we get
$|w^{3}+z^{3}| = \left|\frac{3}{2}(w^{2}+z^{2})-\frac{1}{2}(w+z)^{2}\right| \geq \frac{3}{2}|w^{2}+z^{2}| - \frac{1}{2}|w+z|^{2} = \frac{3}{2}(14) - \frac{1}{2}(1) = \frac{41}{2}.$
To achieve $\frac{41}{2}$ , it suffices to take $w, z$ satisfying $w+z=1$ and $w^{2}+z^{2}=14$ .

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