<s>[INST] Let $a$ and $b$ be two complex numbers such that $|a+b|=20$ and $|a^2+b^2|=16$ then find maximum and minimum value of $|a^3+b^3|$ Let $a$ and $b$ be two complex numbers such that
$|a+b|=20$ and
$|a^2+b^2|=16$
then find maximum and minimum value of $|a^3+b^3|$
My Attempt:
$|a^3+b^3|=|(a+b)((a+b)^2-3ab|)\leq |a+b|^3+3|ab|$
Couldn't go beyond this
 [/INST] We have $|a+b|=20\iff a+b=20e^{it}\iff ae^{-it}+be^{-it}=20$
So WLOG we can replace $a,b$ by phase shifted ones and  assume $a+b=20$.
We also have $|a^2+b^2|=16\iff a^2+b^2=16e^{i\theta}$

$\begin{align}\Big|a^3+b^3\Big|
&=\Big|(a+b)^3-\tfrac 32(\underbrace{(a+b)^2-(a^2+b^2)}_{2ab})(a+b)\Big|\\
&=\Big|20^3-\frac 32((20^2-16e^{i\theta})\times 20\Big|\\\\
&=\Big|480e^{i\theta}-4000\Big|\\\\
&=160\sqrt{634-150\cos(\theta)}\end{align}$

*

*The minimum is when $\cos(\theta)=1\iff \theta=0$ and the value is $3520$
Initial values would be $\begin{cases}a+b=20\\a^2+b^2=16\end{cases}\iff a,b=10\pm 2i\sqrt{23}$

*

*The maximum is when $\cos(\theta)=-1\iff \theta=\pi$ and the value is $4480$
Initial values would be $\begin{cases}a+b=20\\a^2+b^2=-16\end{cases}\iff a,b=10\pm 6i\sqrt{3}$
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