Question

If |z1|=1,|z2|=2,|z3|=3and |9z1z2+4z1z3+z2z3|=12 then value of |z1+z2+z3|=?

Answer 1

$\textbf{Given:}$

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

$\text{consider},$

$|z_1|=1$

$\implies\;|z_1|^2=1^2$

$\implies\;z_1\overline{z_1}=1$

$\implies\;z_1=\frac{1}{\overline{z_1}}$

$\text{Similarly}\;z_2=\frac{4}{\overline{z_2}}\text{ and } z_3=\frac{9}{\overline{z_3}}$

$\text{Now,}$

$|z_1+z_2+z_3|$

$=|\frac{1}{\overline{z_1}}+\frac{4}{\overline{z_2}}+\frac{9}{\overline{z_3}}|$

$=|\frac{\overline{z_2}\overline{z_3}+4\overline{z_1}\overline{z_3}+9\overline{z_1}\overline{z_2}}{\overline{z_1}\overline{z_2}\overline{z_3}}|$

$=|\frac{\overline{z_2z_3+4\,z_1z_3+9\,z_1z_2}}{\overline{z_1z_2z_3}}|$

$=|\overline{(\frac{z_2z_3+4z_1z_3+9z_1z_2}{z_1\,z_2\,z_3})}|$

$\text{We know that, }\bf|z|=|\bar{z}|$

$=|\frac{z_2z_3+4z_1z_3+9z_1z_2}{z_1z_2z_3}|$

$=\frac{|z_2z_3+4z_1z_3+9z_1z_2|}{|z_1|\,|z_2|\,|z_3|}$

$=\frac{12}{1{\times}2{\times}3}$

$=6$

$\therefore\bf|z_1+z_2+z_3|=6$

Answer 2

Answer:

Step-by-step explanation:

That answer is wrong

Answer is 2