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If $z_1$ and $z_2$ are two complex numbers such that $|z_1+z_2|=1$ and $|z_1^2+z_2^2|=25$, find the minimum value of $|z_1^3+z_2^3|$.

My try:

The minimum value is $37$ which I dot by taking $z_1+z_2=1$ and considering only real part and$z_1^2+z_2^2=25$, hence we get $z_1=4$ and $z_2=-3$, solving we get $|z_1^3+z_2^3|=|64-27|=37$

But this is not the appropriate way , it needs to be solved via triangle property or property of complex number, I am not able to solve it via property

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    $\begingroup$ A little observation: $$|z_1^2 + z_2^2| = |(z_1+z_2)^2 - 2z_1z_2| \le |z_1 + z_2|^2 + 2|z_1z_2|$$ This implies $$|z_1z_2| \ge 12$$ $\endgroup$
    – VIVID
    Oct 24, 2021 at 4:38
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    $\begingroup$ Also use $|z_1^3+z_2^3|$ = $|z_1+z_2||z_1^2+z_2^2 - z_1z_2|$ $\endgroup$
    – eyllanesc
    Oct 24, 2021 at 4:50
  • $\begingroup$ Nice, @eyllanesc and vivid. Hence,$$|z_1^3+z_2^3| = |(z_1+z_2)^2 - 3z_1z_2|\ge 3|z_1z_2| - |z_1+z_2|^2\ge 36-1 = 35.$$That's maybe not the solution, but it's getting close. In addition,$$25 = |2z_1z_2 - (z_1+z_2)^2|\ge 2|z_1z_2| - 1,$$ thus $|z_1z_2|\le 13$, so that $|z_1^3+z_2^3|\le 38$. $\endgroup$
    – amsmath
    Oct 24, 2021 at 5:35
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    $\begingroup$ @eyllanesc $|z_1z_2| \ne 12$ but $\ge 12$ $\endgroup$
    – VIVID
    Oct 24, 2021 at 8:37
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    $\begingroup$ $$|z_1^3+z_2^3| = |z_1^2-z_1z_2 + z_2^2| = |\frac32(z_1^2+z_2^2)-\frac12(z_1+z_2)^2|\\ \ge \frac32|z_1^2+z_2^2|-\frac12|z_1+z_2|^2 = 37$$ $\endgroup$ Oct 24, 2021 at 10:08

1 Answer 1

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This answer is made of achille hui's comment.


Observe that $$\begin{align} |z_1^3+z_2^3| &= |z_1^2-z_1z_2 + z_2^2| \\ &= \left|\frac32(z_1^2+z_2^2)-\frac12(z_1+z_2)^2\right| \\ &\ge \frac32|z_1^2+z_2^2|-\frac12|z_1+z_2|^2 \\ &= 37 \end{align}$$ Also, note that the equality holds for $(z_1,z_2) = (4,-3)$. So, indeed, the minimum value of the required expression is $37$.

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