Question

If $\left|z_1\right|=1,\left|z_2\right|=2,\left|z_3\right|=3$ and $|9z_1{z}_2+4z_1{z}_3+z_2{z}_3|=12$ then the value of $|z_1+z_2+z_3|$ is equal to

• A. $2$
• B. $3$
• C. $4$
• D. $6$

Answer 1

The correct option is A $2$

$z\overline{z}={\left|z\right|}^2$

$\overline{z_1+z_2+z_3}=\overline{z_1}+\overline{z_2}+\overline{z_3}$

${|z_1+z_2+z_3|}^2=(z_1+z_2+z_3)(\overline{z_1}+\overline{z_2}+\overline{z_3})$

As $\left|z_3\right|=3$ we have $z_3\overline{z_3}=9$

$z_2\overline{z_2}=4$ and $z_1\overline{z_1}=1$

Now, we can write $|9z_1{z}_2+4z_1{z}_3+z_2{z}_3|=12$

as $|z_3\overline{z_3}{z}_1{z}_2+z_2\overline{z_2}{z}_1{z}_3+z_1\overline{z_1}{z}_2{z}_3|=12$

$\Rightarrow \left|z_1{z}_2{z}_3\right||\overline{z_1}+\overline{z_2}+\overline{z_3}|=12$

$\Rightarrow |\overline{z_1}+\overline{z_2}+\overline{z_3}|=\frac{12}{\left|z_1{z}_2{z}_3\right|}=\frac{12}{1\times 2\times 3}=2$