Maths Olympiad Prep

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Algebra Difficulty 5.3 AIME, harder Find the answer

Given that ww and zz are complex numbers such that ∣w+z∣=1|w+z|=1 and ∣w2+z2∣=14\left|w^{2}+z^{2}\right|=14, find the smallest possible value of ∣w3+z3∣\left|w^{3}+z^{3}\right|. Here, ∣⋅∣|\cdot| denotes the absolute value of a complex number, given by ∣a+bi∣=a2+b2|a+b i|=\sqrt{a^{2}+b^{2}} whenever aa and bb are real numbers.

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

We can rewrite ∣w3+z3∣=∣w+z∣∣w2−wz+z2∣=∣w2−wz+z2∣=∣32(w2+z2)−12(w+z)2∣\left|w^{3}+z^{3}\right|=|w+z|\left|w^{2}-w z+z^{2}\right|=\left|w^{2}-w z+z^{2}\right|=\left|\frac{3}{2}\left(w^{2}+z^{2}\right)-\frac{1}{2}(w+z)^{2}\right|Bythetriangleinequality, By the triangle inequality, ∣32(w2+z2)−12(w+z)2+12(w+z)2∣≤∣32(w2+z2)−12(w+z)2∣+∣12(w+z)2∣\left|\frac{3}{2}\left(w^{2}+z^{2}\right)-\frac{1}{2}(w+z)^{2}+\frac{1}{2}(w+z)^{2}\right| \leq\left|\frac{3}{2}\left(w^{2}+z^{2}\right)-\frac{1}{2}(w+z)^{2}\right|+\left|\frac{1}{2}(w+z)^{2}\right|.Byrearrangingandsimplifying,weget. By rearranging and simplifying, we get ∣w3+z3∣=∣32(w2+z2)−12(w+z)2∣≥32∣w2+z2∣−12∣w+z∣2=32(14)−12=412\left|w^{3}+z^{3}\right|=\left|\frac{3}{2}\left(w^{2}+z^{2}\right)-\frac{1}{2}(w+z)^{2}\right| \geq \frac{3}{2}\left|w^{2}+z^{2}\right|-\frac{1}{2}|w+z|^{2}=\frac{3}{2}(14)-\frac{1}{2}=\frac{41}{2}.Toachieve. To achieve 41 / 2,itsufficestotake, it suffices to take w, zsatisfying satisfying w+z=1and and w^{2}+z^{2}=14$.

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