Question

If $\left|z_1\right|=2,\left|z_2\right|=3,\left|z_3\right|=4$ and $|{2z}_1+3z_2+{4z}_3|=4$ then the absolute value of ${8z}_3{z}_2+{27z}_3{z}_1+{64z}_1{z}_2$ equals

• A. $24$
• B. $48$
• C. $72$
• D. $96$

Answer 1

The correct option is D $96$

Given $|z_1|=2,|z_2|=3,|z_3|=4,|2z_1+3z_2+4z_3|=4$

$|8z_3{z}_2+27z_3{z}_1+64z_1{z}_2|=\left|\right(z_1{z}_2{z}_3\left)(\frac{8}{z_1}+\frac{27}{z_2}+\frac{64}{z_3})\right|$

$=|z_1{z}_2{z}_3||\frac{8}{z_1}+\frac{27}{z_2}+\frac{64}{z_3}|$

$=|z_1{z}_2{z}_3||\frac{8\overline{z_1}}{z_1\overline{z_1}}+\frac{27\overline{z_2}}{z_2\overline{z_2}}+\frac{64\overline{z_3}}{z_3\overline{z_3}}|$

$=|z_1||z_2||z_3||\frac{8\overline{z_1}}{|z_1{|}^2}+\frac{27\overline{z_2}}{|z_2{|}^2}+\frac{64\overline{z_3}}{|z_3{|}^2}|$

$[\because z_1\overline{z_1}=|z_1{|}^2]$

$=\mathrm{2.3.4}|\frac{8\overline{z_1}}{4}+\frac{27\overline{z_2}}{9}+\frac{64\overline{z_3}}{16}|$

$=24|2\overline{z_1}+3\overline{z_2}+4\overline{z_3}|$

$=24|\overline{2z_1+3z_2+4z_3}|$

$[\because \overline{z_1+z_2}=\overline{z_1}+\overline{z_2}]$

$=24|2z_1+3z_2+4z_3|$ $[\because |z|=|\overline{z}|]$

$=24.4$

$=96$.