Question

Let $z_K$ ; K = 1, 2, 3, 4 be four complex numbers such that $|z_k|=\sqrt{K+1}\&|30z_1+20z_2+15z_3+12z_4|=K|\sum z_1{z}_2{z}_3|$ then k =

• A. $|z_1{z}_2{z}_4|$
• B. $|z_2{z}_3{z}_4|$
• C. $|z_1{z}_3{z}_4|$
• D. $1$

Answer 1

The correct option is A $|z_1{z}_2{z}_4|$

$\left|z_k\right|=\sqrt{k+1}$

$\left|z_1\right|=\sqrt{2},\left|z_2\right|=\sqrt{3},\sqrt{4}=\left|z_3\right|,\left|z_4\right|=\sqrt{5}$

$\to |30z_1+20z_2+15z_3+12z_4|$

$\to |\frac{30{\left|z_1\right|}^2}{\overline{z_1}}+\frac{20{\left|z_2\right|}^2}{\overline{z_2}}+\frac{15{\left|z_3\right|}^2}{\overline{z_3}}+\frac{12{\left|z_4\right|}^2}{\overline{z_4}}|$

$\to |\frac{30\times 2}{\overline{z_1}}+\frac{20\times 3}{\overline{z_2}}+\frac{15\times 4}{\overline{z_3}}+\frac{12\times 5}{\overline{z_4}}|$

$\to \left(60\right)|\frac{1}{\overline{z_1}}+\frac{1}{\overline{z_2}}+\frac{1}{\overline{z_3}}+\frac{1}{\overline{z_4}}|$

$\to \left(60\right)\left|\overline{\frac{1}{z_1}+\frac{1}{z_2}+\frac{1}{z_3}+\frac{1}{z_4}}\right|=\left(60\right)|\frac{1}{z_1}+\frac{1}{z_2}+\frac{1}{z_3}+\frac{1}{z_4}|$

$\to \left(60\right)\left|\frac{\sum z_1{z}_2{z}_3}{z_1{z}_2{z}_3{z}_4}\right|=\frac{\left(60|\sum z_1{z}_2{z}_3|\right)}{\left|z_1\right|\left|z_2\right|\left|z_3\right|\left|z_4\right|}$

$\to \frac{60\times |\sum z_1{z}_2{z}_3|}{\sqrt{2}\times \sqrt{3}\sqrt{4}\sqrt{5}}$

$\Rightarrow \sqrt{30}|\sum z_1{z}_2{z}_3|=k|\sum z_1{z}_2{z}_3|$

$k=\sqrt{30}=\sqrt{2}\times \sqrt{3}\times \sqrt{5}=\left|z_1\right|\left|z_2\right|\left|z_4\right|$

$k=\left|z_1{z}_2{z}_4\right|$