Question

If $|z_1|=1,|z_2|=2,|z_3|=3$ and $|z_1+2z_2+3z_3|=6,$ then the value of $\frac{1}{6}|z_2{z}_3+8z_1{z}_3+27z_1{z}_2|$ is equal to

Answer 1

We have
$\frac{1}{6}|z_2{z}_3+8z_1{z}_3+27z_1{z}_2|=\frac{|z_1||z_2||z_3|}{6}|\frac{1}{z_1}+\frac{8}{z_2}+\frac{27}{z_3}|$
$\because z_1{\overline{z}}_1=|z_1{|}^2=1\Rightarrow {\overline{z}}_1=\frac{1}{z_1}$
Similarly ${\overline{z}}_2=\frac{4}{z_2}$ and ${\overline{z}}_3=\frac{9}{z_3}$
$\therefore \frac{|z_1||z_2||z_3|}{6}|{\overline{z}}_1+2{\overline{z}}_2+3{\overline{z}}_3|$
$=\frac{|z_1||z_2||z_3|}{6}\left|\overline{z_1+2z_2+3z_3}\right|(\because |z|=|\overline{z}|$
$=\frac{|z_1||z_2||z_3|}{6}|z_1+2z_2+3z_3|$
$=\frac{1\times 2\times 3}{6}\times 6=6$