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Given that $|z_1|=|z_1+z_2|=3$ and $|z_1-z_2|=3\sqrt{3}$, I want check my following solution in determining the value of $\lfloor\log_3{|(z_1\bar{z_2})^{2022}+(\bar{z_1}z_2)^{2022}|}\rfloor$.

My solution: I started by using $|z_1+z_2|^2=3^2$ which gave $(z_1+z_2)(\bar{z_1}+\bar{z_2})=9$
$|z_1|^2+z_2\bar{z_1}+z_1\bar{z_2}+|z_2|^2=9$.

Squaring both sides of $|z_1-z_2|=3\sqrt{3}$ I get $|z_1|^2-z_2\bar{z_1}-z_1\bar{z_2}+|z_2|^2=27$.

Adding these two equations and subbing in $|z_1|=3$ gives $18+2|z_2|^2=27+9\rightarrow |z_2|=3$.

Substituting this back into either equation gives $z_1\bar{z_2}+\bar{z_1}z_2=-9$ which is $\text{Re}(z_1\bar{z_2})=-\frac{9}{2}$. Now denote $z_1\bar{z_2}=-\frac{9}{2}+iy$.

Knowing that $|z_1|=|z_2|=|\bar{z_2}|=3$, then $|z_1\bar{z_2}|=9$.

$\therefore\left(-\frac{9}{2}\right)^2+y^2=81\rightarrow y=\pm\frac{9\sqrt{3}}{2}$

$z_1\bar{z_2}=9e^{i2\pi/3}$ and $\bar{z_1}z_2=9e^{-i2\pi/3}$ (or other way around it doesn't affect the solution).

$\therefore |(z_1\bar{z_2})^{2022}+(\bar{z_1}z_2)^{2022}|=|9^{2022}\times 2\cos{(1348\pi)}|=|9^{2022}\times2|$

$\therefore \lfloor\log_3{|(z_1\bar{z_2})^{2022}+(\bar{z_1}z_2)^{2022}|}\rfloor=4044$.

As well as checking this solution, I would be happy to see if there are any more elegant ways to complete this.

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    $\begingroup$ Your solution is optimal. $\endgroup$ Dec 5, 2022 at 11:06

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Good afternoon! In my opinion, you can use the geometric interpretation of complex numbers here. z1 - This is a vector whose length is 3 and which moves only along a circle of radius 3 depending on the angle $ \phi $ $$ |z_{1}| = 3 \Rightarrow z_{1} = 3e^{i\phi} , \phi\in \mathbb{R} $$ Now we can think, draw vectors on a piece of paper and understand that the only arrangement of vectors z1 and z2 in which their sum is equal to the length of one of them is the case when the triangle formed by these vectors is correct(or perfect)! (all sides are equal)

From here we get $$ z_{1} = 3e^{i\phi} z_{2} = 3e^{i\psi} ,~~~ \phi,\psi\in \mathbb{R} $$

Also, for geometric reasons, we understand that there must be a connection between the angles $\phi,\psi$. Let's find her!

$$ |3e^{i\phi} + 3e^{i\psi}| = 3, $$ $$ |\cos(\phi)+i\sin(\phi) + (\cos(\psi)+i\sin(\psi))| = 1 ,$$ $$ |(\cos(\phi) + \cos(\psi))+i(\sin(\phi) + \sin(\psi))| = 1 , $$ $$ \sqrt{\cos(\phi) + \cos(\psi))^2 + (\sin(\phi) + \sin(\psi))^2} = 1 ,$$ $$ \sqrt{2 + 2\cos(\phi - \psi)} = 1 , $$ and finally we get $$ \cos( \frac{\phi - \psi}{2}) = \frac{1}{2} \Rightarrow \frac{\phi - \psi}{2} = \pm \frac{\pi}{3} + 2\pi k, ~~k\in \mathbb {Z}$$ For certainty , let 's take the case. $$ \phi - \psi = \frac{2\pi}{3} $$

That's all, it remains only to substitute the found values into the desired expression $$ z_{1} = 3e^{i\phi}, z_{2} = 3e^{i(\phi - \frac{2\pi}{3}) } $$ $$ \overline z_{1} = 3e^{-i\phi} , \overline z_{2} =3e^{-i(\phi - \frac{2\pi}{3}) } $$

$$ (z_{1} \overline z_{2})^{2022} = 3^{4044}\cdot e^{674i}$$ $$ ( \overline z_{1} z_{2})^{2022} = 3^{4044}\cdot e^{-674i}$$ $$ |(z_{1} \overline z_{2})^{2022} + ( \overline z_{1} z_{2})^{2022}| = 3^{4044}$$

$$ \log_{3} 3^{4044} = 4044 $$ $$ \lfloor 4044 \rfloor = 4044 $$ Answer: 4044

P.S. Let's make sure that the equation below is correct $$ |z_{1} - z_{2}| = 3\sqrt{3}$$ $$ 3|e^{i\phi} - e^{i(\phi - \frac{2\pi}{3})}| = 3\sqrt{3} , ~~|e^{i\phi}| = 1$$ $$ |e^{i\phi} - e^{i(\phi - \frac{2\pi}{3})}| = \sqrt{3} \cdot e^{\Im(\phi)} = \sqrt{3}$$ (sorry for my english :))

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