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Given $w$ and $z$ two complex numbers such that $|z+w| = 1$ and $|z^2+w^2| = 14$. Find the smallest possible value of $|w^3+z^3|$, where |.| denotes the absolute value of a complex number, given by $|a+bi| = \sqrt{a^2 + b^2}$.

My working
$|w^3+z^3| = |w+z||w^2+z^2-zw| = |(w+z)^2-3wz|$
Using triangle inequalities,
$|(w+z)^2-3wz| \ge ||w+z|^2-3|wz|| = |1-3|wz||$
$|z+w|^2 = |z^2+w^2+2wz|$
again applying triangle inequality
$|z^2+w^2|+2|wz|\ge1\ge||w^2+z^2|-2|wz||$
solving for range of |wz| I get
$15/2\ge |wz|\ge13/2$ then i getting the incorrect range of of $|z^3+w^2|$.

user26857
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Chhaya gupta
  • 33
  • 6

2 Answers2

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We can write the given conditions as $z+w=\mathrm e^{\mathrm i\theta}$ and $z^2+w^2=14\mathrm e^{\mathrm i\phi},$ for some angles $\theta$ and $\phi$. Now $$\begin{align}z^3 +w^3&=\tfrac12(z+w)[3(z^2+w^2)-(z+w)^2]\\&=21\mathrm e^{\mathrm i(\theta+\phi)}-\tfrac12\mathrm e^{3\mathrm i\theta}\\&=21\cos(\theta+\phi)-\tfrac12\cos3\theta+\mathrm i[21\sin(\theta +\phi)-\tfrac12\sin3\theta].\end{align}$$ Therefore $$\begin{align}|z^3+w^3|^2&=[21\cos(\theta+\phi)-\tfrac12\cos3\theta]^2+[21\sin(\theta +\phi)-\tfrac12\sin3\theta]^2\\&=441.25-21\cos(\phi-2\theta).\end{align}$$ This is minimized when $\cos(\phi-2\theta)=1$, namely when $$|z^3+w^3|=\surd420.25=20.5.$$

John Bentin
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-1

Mathematica 14 answers

z = x + I*y; w = u + I*v; NMinimize[ComplexExpand[{Abs[w^3 + z^3]^2, 
Abs[z + w]^2 == 1 && Abs[z^2 + w^2]^2 == 14^2}], {x, y, u, v}]

$$\{420.25,\{x -> -2.05877, y -> -0.404213, u -> 3.04004, v -> 0.596871\}\} $$

I have strong doubts concerning its symbolic solution.

user64494
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