Question

Let $|{z_i}| = i,i = 1,2,3,4$and $|16{z_1}{z_2}{z_3} + 9{z_1}{z_2}{z_4} + 4{z_1}{z_3}{z_4} + {z_2}{z_3}{z_4}| = 48$, then the value of $|\dfrac{1}{{{z_1}}} + \dfrac{4}{{{z_2}}} + \dfrac{9}{{{z_3}}} + \dfrac{{16}}{{{z_4}}}|$.
(A). 1
(B). 2
(C). 4
(D). 8

Answer 1

Hint – We know, $|{z_i}| = i,i = 1,2,3,4$, so, $|{z_1}| = 1$, $|{z_2}| = 2$, $|{z_3}| = 3$, $|{z_4}| = 4$, Now take out $|{z_1}||{z_2}||{z_3}||{z_4}|$ common from the $2_{nd}$ equation to obtain the required expression.

Complete step-by-step answer:
According to question we have,
$|{z_i}| = i,i = 1,2,3,4$
Therefore, $|{z_1}| = 1$, $|{z_2}| = 2$, $|{z_3}| = 3$, $|{z_4}| = 4$.
Also, $|16{z_1}{z_2}{z_3} + 9{z_1}{z_2}{z_4} + 4{z_1}{z_3}{z_4} + {z_2}{z_3}{z_4}| = 48$
Solving the given equation further by taking $|{z_1}||{z_2}||{z_3}||{z_4}|$ common, we get-
\[
  |16{z_1}{z_2}{z_3} + 9{z_1}{z_2}{z_4} + 4{z_1}{z_3}{z_4} + {z_2}{z_3}{z_4}| = 48 \\
   = |{z_1}||{z_2}||{z_3}||{z_4}||\dfrac{{16}}{{{z_4}}} + \dfrac{9}{{{z_3}}} + \dfrac{4}{{{z_2}}} + \dfrac{1}{{{z_1}}}| = 48 - (1) \\
 \]
Put the value of $|{z_1}| = 1$, $|{z_2}| = 2$, $|{z_3}| = 3$, $|{z_4}| = 4$ in equation (1), we get-
\[
   = (1)(2)(3)(4)|\dfrac{{16}}{{{z_4}}} + \dfrac{9}{{{z_3}}} + \dfrac{4}{{{z_2}}} + \dfrac{1}{{{z_1}}}| = 48 \\
   \Rightarrow |\dfrac{1}{{{z_1}}} + \dfrac{4}{{{z_2}}} + \dfrac{9}{{{z_3}}} + \dfrac{{16}}{{{z_4}}}| = \dfrac{{48}}{{24}} = 2 \\
 \]
Hence, the answer is option (B), i.e. \[|\dfrac{1}{{{z_1}}} + \dfrac{4}{{{z_2}}} + \dfrac{9}{{{z_3}}} + \dfrac{{16}}{{{z_4}}}| = 2\].

Note – Whenever such types of questions appear, then always write the information given in the question. Use that information to form equations and solve the question. As, mentioned in the solution, we have found out the values of $|{z_1}| = 1$, $|{z_2}| = 2$, $|{z_3}| = 3$, $|{z_4}| = 4$ by using Let $|{z_i}| = i,i = 1,2,3,4$, which is provided in the question and then using these values, we have solved the question.