Question

If ${m_1},{m_2},{m_3},{m_4}$ denote the moduli of the complex numbers $1 + 4i,3 + i,1 - i,2 - 3i$, then the correct one among the following is
$1){m_1} < {m_2} < {m_3} < {m_4}$
$2){m_4} < {m_3} < {m_2} < {m_1}$
$3){m_3} < {m_2} < {m_4} < {m_1}$
$4){m_3} < {m_1} < {m_2} < {m_4}$

Answer 1

Hint: First, complex numbers are the real and imaginary combined numbers as in the form of $z = x + iy$, where x and y are the real numbers and $i$ is the imaginary.
Imaginary $i$ can be also represented into the real values only if, ${i^2} = - 1$
The modulus of the complex number can be expressed as $\left| {x + iy} \right|$
Formula used:
The modulus of the complex number denoted in the square root as $\left| {x + iy} \right| = \sqrt {{x^2} + {y^2}} $

Complete step-by-step solution:
Since from the given that we have, ${m_1},{m_2},{m_3},{m_4}$ denote the moduli of the complex numbers $1 + 4i,3 + i,1 - i,2 - 3i$.
Let us write the given in the mathematical expression, ${m_1} = \left| {1 + 4i} \right|,{m_2} = \left| {3 + i} \right|,{m_3} = \left| {1 - i} \right|,{m_4} = \left| {2 - 3i} \right|$
From the given formula, the modulus of the complex number denoted in the square root as $\left| {x + iy} \right| = \sqrt {{x^2} + {y^2}} $
First, take the value ${m_1} = \left| {1 + 4i} \right|$ then expressed this into the square root, we get ${m_1} = \left| {1 + 4i} \right| \Rightarrow \sqrt {{1^2} + {4^2}} $ where $x = 1,y = 4$
Thus, solving this we get ${m_1} = \left| {1 + 4i} \right| \Rightarrow \sqrt {{1^2} + {4^2}} = \sqrt {1 + 16} \Rightarrow \sqrt {17} $
Similarly, take the value ${m_2} = \left| {3 + i} \right|$ then expressed this into the square root, we get ${m_2} = \left| {3 + i} \right| \Rightarrow \sqrt {{3^2} + {1^2}} $ where $x = 3,y = 1$
Thus, solving this we get ${m_2} = \left| {3 + i} \right| \Rightarrow \sqrt {{3^2} + {1^2}} = \sqrt {9 + 1} \Rightarrow \sqrt {10} $
Similarly, take the value ${m_3} = \left| {i - 1} \right|$ then expressed this into the square root, we get ${m_3} = \left| {i - 1} \right| \Rightarrow \sqrt {{1^2} + {{( - 1)}^2}} $ where $x = - 1,y = 1$
Thus, solving this we get ${m_3} = \left| {i - 1} \right| \Rightarrow \sqrt {{1^2} + {{( - 1)}^2}} = \sqrt {1 + 1} \Rightarrow \sqrt 2 $
Similarly, take the value ${m_4} = \left| {2 - 3i} \right|$ then expressed this into the square root, we get ${m_4} = \left| {2 - 3i} \right| \Rightarrow \sqrt {{2^2} + {{( - 3)}^2}} $ where $x = 2,y = - 3$
Thus, solving this we get ${m_4} = \left| {2 - 3i} \right| \Rightarrow \sqrt {{2^2} + {{( - 3)}^2}} = \sqrt {4 + 9} \Rightarrow \sqrt {13} $
Hence, we get the relation, $\sqrt {17} > \sqrt {13} > \sqrt {10} > \sqrt 2 $ which can be also represented as $\sqrt {17} > \sqrt {13} > \sqrt {10} > \sqrt 2 \Rightarrow \sqrt 2 < \sqrt {10} < \sqrt {13} < \sqrt {17} = {m_3} < {m_2} < {m_4} < {m_1}$
Therefore, the option $3){m_3} < {m_2} < {m_4} < {m_1}$ is correct.

Note: Since in the algebraic concept, we know that $a < b = b > a$ because substitute the value of $b = 2,a = 1$ then we get the relation as $2 > 1 = 1 < 2$ and hence less than the reverse process is greater than and we applied this concept in $\sqrt {17} > \sqrt {13} > \sqrt {10} > \sqrt 2 \Rightarrow \sqrt 2 < \sqrt {10} < \sqrt {13} < \sqrt {17} $.
We were also able to find the relation using the square root concept, which is $\sqrt 2 = 1.414$ and after finding every value of the root terms, compare and we get the same result as above.