For three complex numbers we have:

$|z_1|=1$ ,$|z_2|=2$ ,$|z_3|=3$


$|9z​_1z_2 + 4z_1z_3 + z_2z_3|=12$

Then find value of $|z_1 + z_2 + z_3|$

I took $z_1=1(\cos A+i\sin A),z_2=2(\cos B+i\sin B),z_3=3(\cos A+i\sin A)$ but it doesn't help much. Any hint?

What I could see is that $|9z​_1z_2 + 4z_1z_3 + z_2z_3|=12$ can be rewritten as

$||z_3|^2z​_1z_2 +|z_2|^2z_1z_3 + |z_1|^2 z_2z_3|=12$ , but not so sure if it is useful.


As $|z_3|=3$ we have $z_3 \bar{z_3}=9$ and similarly $z_2 \bar{z_2}=4$ and $z_1 \bar{z_1}=1$

Now we can write $$|9z_1z_2+4z_1z_3+z_2z_3|=12$$ as

$$|\bar{z_3}z_3 z_1 z_2+\bar{z_2} z_2 z_1z_3+\bar{z_1}z_1z_2z_3|=12$$ $\implies$

$$|z_1z_2z_3||\bar{z_1}+\bar{z_2}+\bar{z_3}|=12$$ so

$$|{z_1+z_2+z_3}|=\frac{12}{1 \times 2 \times 3}=2$$

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  • $\begingroup$ Very nice solution! $\endgroup$ – MrYouMath Apr 8 '16 at 7:41

You can have $$|9z_1z_2+4z_1z_3+z_2z_3|^2=12^2,$$ i.e. $$(9z_1z_2+4z_1z_3+z_2z_3)(9\overline{z_1}\overline{z_2}+4\overline{z_1}\overline{z_3}+\overline{z_2}\overline{z_3})=12^2$$ Expanding the LHS, $$9^2|z_1|^2|z_2|^2+4^2|z_1|^2|z_3|^2+|z_2|^2|z_3|^2+36|z_1|^2(z_2\overline{z_3}+\overline{z_2}z_3)+9|z_2|^2(z_1\overline{z_3}+\overline{z_1}z_3)+4|z_3|^2(z_1\overline{z_2}+\overline{z_1}z_2)=144$$ and so $$z_2\overline{z_3}+\overline{z_2}z_3+z_1\overline{z_3}+\overline{z_1}z_3+z_1\overline{z_2}+\overline{z_1}z_2=-10$$

Using this, $$\begin{align}&|z_1+z_2+z_3|^2\\&=|z_1|^2+|z_2|^2+|z_3|^2+z_2\overline{z_3}+\overline{z_2}z_3+z_1\overline{z_3}+\overline{z_1}z_3+z_1\overline{z_2}+\overline{z_1}z_2\\&=1+4+9-10\\&=4\end{align}$$ and so $$|z_1+z_2+z_3|=\color{red}{2}.$$

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