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I'm kind of stuck at this. My progress so far:

Suppose $w,z\in\mathbb{C}$ are such that $|w+z|=10$ and $|w^2+z^2|=11$. Let $a=z+w$ and $b=z^2+w^2$. One can observe that $\frac{a^2-b}{2}=zw$, so we have:

$|w^3+z^3|=|w+z||w^2+z^2-zw|=10|b-\frac{a^2-b}{2}|=5|3b-a^2|\geq 5||3b|-|a^2||=5|3|11|-100|=5\cdot 67=335$

so $335$ is a lower bound for all the possible values of $|w^3+z^3|$ given such conditions. If I could show that there acutally exist a couple of complex numbers that satisfy $|3b-a^2|= ||3b|-|a^2||$ then I would have that $|w^3+z^3|=335$ and I would've found the minimum, since I already showed $335$ is a lower bound.

Since there are so much requirements involved, the algebra is just such a mess and I don't even know where I could start searching for those two numbers.

Any help is greatly appreciated. Greetings.

Bryan Castro
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    No idea on why but $w=5+ki,z=5-ki$ solving for $k$ you actually get $|w^3+z^3|=335$, maybe has something to do with $z=\overline{w} $ – kingW3 Oct 03 '21 at 21:50
  • Wouldn’t 3b and a^2 just need to be collinear by the triangle inequality? – Average Oct 03 '21 at 22:53
  • @newToProgramming yes indeed but at least I could'nt do anything with that fact. – Bryan Castro Oct 03 '21 at 22:55
  • @kingW3 I have no Idea how you came up with that, but in fact $z=5+\sqrt{\frac{39}{2}}i$ and $w=5-\sqrt{\frac{32}{9}}i$ work. What's crazy is, I solved for $k$ trying to force $|z^2+w^2|=11$, and when I calculated $|z^3+w^3|$ it was $335$. Thank you so much. – Bryan Castro Oct 03 '21 at 22:58
  • You just force $z=\bar w$. This makes $a$ and $b$ colinear, which gets you the equality. – Trebor Oct 04 '21 at 00:33

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