If $m_{1}, m_{2}, m_{3}$ and $m_{4}$ respectively denote the moduli of the complex numbers $1 + 4 i, 3 + i, 1 - i$ and $2 - 3 i$ , then the correct one, among the following is:

$m_{1} < m_{2} < m_{3} < m_{4}$

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$m_{4} < m_{3} < m_{2} < m_{1}$

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$m_{3} < m_{2} < m_{4} < m_{1}$

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$m_{3} < m_{1} < m_{2} < m_{4}$

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Solution

The correct option is C
$m_{3} < m_{2} < m_{4} < m_{1}$

Explanation for the correct option:
Modulus of complaex number :
We know that modulus of a complex number $z = x i + y$ is $\sqrt{x^{2} + y^{2}}$
So, the modulus of given complex numbers will be,
$\Rightarrow |m_{1}| = |1 + 4 i| = \sqrt{4^{2} + 1^{2}} = \sqrt{17} \Rightarrow |m_{2}| = |3 + i| = \sqrt{1^{2} + 3^{2}} = \sqrt{10} \Rightarrow |m_{3}| = |1 - i| = \sqrt{{(- 1)}^{2} + 1^{2}} = \sqrt{2} \Rightarrow |m_{4}| = |2 - 3 i| = \sqrt{{(- 3)}^{2} + 2^{2}} = \sqrt{13}$
So, $m_{3} < m_{2} < m_{4} < m_{1}$
Thus, option (C) $m_{3} < m_{2} < m_{4} < m_{1}$ is the correct answer.

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