Question

Let $z_r(i\le r\le 4)$ be complex numbers such that $|Z_r|=\sqrt{r+1}$ and $|30z_1+20z_2+15z_3+12z_4|=K|z_1{z}_2{z}_3+z_2{z}_3{z}_4+z_3{z}_4{z}_1+z_4{z}_1{z}_2|$. Then the value of k equals

• A. $|z_1{z}_2{z}_3|$
• B. $|z_2{z}_3{z}_4|$
• C. $|z_3{z}_4{z}_1|$
• D. $|z_4{z}_1{z}_2|$

Answer 1

Answer: (A)

$|z_1{z}_2{z}_3|$

$|z_r|=\sqrt{r+1}\therefore |z_1|=\sqrt{2},|z_2|=\sqrt{3},|z_3|=2$ & $|z_4|=\sqrt{5}$
Now $|30z_1+20z_2+15z_3+12z_4|=k|z_1{z}_2{z}_3+z_2{z}_3{z}_4+.....|$
LHS $=|30\frac{2}{z_1}+20\frac{3}{z_2}+15\frac{4}{z_3}+12\frac{5}{z_4}|$
$|z_1{|}^2=2$ $|z_2|=3$, $|z_3|=4$, $|z_4{|}^2=5$
$z_1\overline{z_1}=2z_2\overline{z_2}=3z_3\overline{z_3}=4z_4\overline{z_4}=5$
$z_1=\frac{2}{\overline{z_1}},z_2=\frac{3}{\overline{z_2}},z_4=\frac{5}{\overline{z_4}}$
$LHS=60\left|\frac{\overline{z_1{z}_2{z}_3}+\overline{z_2{z}_3{z}_4}+...+....}{z_1{z}_2{z}_3{z}_4}\right|$
$=60\left|\frac{z_1{z}_2{z}_3+z_2{z}_3{z}_4+...}{z_1{z}_2{z}_3{z}_4}\right|$
$k=\frac{0}{|z_1{z}_2{z}_3{z}_4|}=\frac{2\times 3\times 5\times 2}{|z_1{z}_2{z}_3{z}_4|}$
$=\frac{|z_1{|}^2|z_2{|}^2|z_4{|}^2|z_3|}{|z_1{z}_2{z}_4{z}_3|}$
$=|z_1{z}_2{z}_4|$