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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\...

If m1,m2,m3,m4{m_1},{m_2},{m_3},{m_4} denote the moduli of the complex numbers 1+4i,3+i,1−i,2−3i1 + 4i,3 + i,1 - i,2 - 3i, then the correct one among the following is
1)m1<m2<m3<m41){m_1} < {m_2} < {m_3} < {m_4}
2)m4<m3<m2<m12){m_4} < {m_3} < {m_2} < {m_1}
3)m3<m2<m4<m13){m_3} < {m_2} < {m_4} < {m_1}
4)m3<m1<m2<m44){m_3} < {m_1} < {m_2} < {m_4}

Explanation

Solution

First, complex numbers are the real and imaginary combined numbers as in the form of z=x+iyz = x + iy, where x and y are the real numbers and ii is the imaginary.
Imaginary ii can be also represented into the real values only if, i2=−1{i^2} = - 1
The modulus of the complex number can be expressed as ∣x+iy∣\left| {x + iy} \right|
Formula used:
The modulus of the complex number denoted in the square root as ∣x+iy∣=x2+y2\left| {x + iy} \right| = \sqrt {{x^2} + {y^2}}

Complete step-by-step solution:
Since from the given that we have, m1,m2,m3,m4{m_1},{m_2},{m_3},{m_4} denote the moduli of the complex numbers 1+4i,3+i,1−i,2−3i1 + 4i,3 + i,1 - i,2 - 3i.
Let us write the given in the mathematical expression, m1=∣1+4i∣,m2=∣3+i∣,m3=∣1−i∣,m4=∣2−3i∣{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 ∣x+iy∣=x2+y2\left| {x + iy} \right| = \sqrt {{x^2} + {y^2}}
First, take the value m1=∣1+4i∣{m_1} = \left| {1 + 4i} \right| then expressed this into the square root, we get m1=∣1+4i∣⇒12+42{m_1} = \left| {1 + 4i} \right| \Rightarrow \sqrt {{1^2} + {4^2}} where x=1,y=4x = 1,y = 4
Thus, solving this we get m1=∣1+4i∣⇒12+42=1+16⇒17{m_1} = \left| {1 + 4i} \right| \Rightarrow \sqrt {{1^2} + {4^2}} = \sqrt {1 + 16} \Rightarrow \sqrt {17}
Similarly, take the value m2=∣3+i∣{m_2} = \left| {3 + i} \right| then expressed this into the square root, we get m2=∣3+i∣⇒32+12{m_2} = \left| {3 + i} \right| \Rightarrow \sqrt {{3^2} + {1^2}} where x=3,y=1x = 3,y = 1
Thus, solving this we get m2=∣3+i∣⇒32+12=9+1⇒10{m_2} = \left| {3 + i} \right| \Rightarrow \sqrt {{3^2} + {1^2}} = \sqrt {9 + 1} \Rightarrow \sqrt {10}
Similarly, take the value m3=∣i−1∣{m_3} = \left| {i - 1} \right| then expressed this into the square root, we get m3=∣i−1∣⇒12+(−1)2{m_3} = \left| {i - 1} \right| \Rightarrow \sqrt {{1^2} + {{( - 1)}^2}} where x=−1,y=1x = - 1,y = 1
Thus, solving this we get m3=∣i−1∣⇒12+(−1)2=1+1⇒2{m_3} = \left| {i - 1} \right| \Rightarrow \sqrt {{1^2} + {{( - 1)}^2}} = \sqrt {1 + 1} \Rightarrow \sqrt 2
Similarly, take the value m4=∣2−3i∣{m_4} = \left| {2 - 3i} \right| then expressed this into the square root, we get m4=∣2−3i∣⇒22+(−3)2{m_4} = \left| {2 - 3i} \right| \Rightarrow \sqrt {{2^2} + {{( - 3)}^2}} where x=2,y=−3x = 2,y = - 3
Thus, solving this we get m4=∣2−3i∣⇒22+(−3)2=4+9⇒13{m_4} = \left| {2 - 3i} \right| \Rightarrow \sqrt {{2^2} + {{( - 3)}^2}} = \sqrt {4 + 9} \Rightarrow \sqrt {13}
Hence, we get the relation, 17>13>10>2\sqrt {17} > \sqrt {13} > \sqrt {10} > \sqrt 2 which can be also represented as 17>13>10>2⇒2<10<13<17=m3<m2<m4<m1\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)m3<m2<m4<m13){m_3} < {m_2} < {m_4} < {m_1} is correct. **

Note: Since in the algebraic concept, we know that a<b=b>aa < b = b > a because substitute the value of b=2,a=1b = 2,a = 1 then we get the relation as 2>1=1<22 > 1 = 1 < 2 and hence less than the reverse process is greater than and we applied this concept in 17>13>10>2⇒2<10<13<17\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 2=1.414\sqrt 2 = 1.414 and after finding every value of the root terms, compare and we get the same result as above.