Question

- If \[\dfrac{{5{z_2}}}{{11{z_1}}}\] is purely imaginary , then find the value of \[\left| {\dfrac{{\left( {2{z_1} + 3{z_2}} \right)}}{{\left( {2{z_1} - 3{z_2}} \right)}}} \right|\]
A. \[\dfrac{{37}}{{33}}\]
B. \[2\]
C. \[1\]
D. \[3\]
E. \[\dfrac{{33}}{{37}}\]

Answer 1

Hint: We will solve the question with the property of complex numbers followed by the property of conjugate of the complex number after that we will find the modulus of the complex number.

Formula Used:-
Property Of conjugate of Complex number will be used in this question which is If \[z = a + ib\] then its conjugate is \[z = a - ib\] and modulus of a complex number is equal to the modulus of its conjugate which is given by \[\left| z \right| = \sqrt {{a^2} + {b^2}} \].

Complete step by step solution:
According to the question its given that \[\dfrac{{5{z_2}}}{{11{z_1}}}\] is purely imaginary which means Real Part of the complex number is \[0\] .
\[z = a + ib\]
Where \[a = 0\] and \[ib = \dfrac{{5{z_2}}}{{11{z_1}}}\]
Now proceeding with this concept,
Let \[ik = \dfrac{{5{z_2}}}{{11{z_1}}}\] where \[k = \] constant and \[i = \sqrt { - 1} \]
\[\dfrac{{5{z_2}}}{{11{z_1}}} = ik\]
\[\dfrac{{{z_2}}}{{{z_1}}} = \dfrac{{11}}{5}ik\]
Now in the given modulus function \[\left| {\dfrac{{\left( {2{z_1} + 3{z_2}} \right)}}{{\left( {2{z_1} - 3{z_2}} \right)}}} \right|\] we will be dividing Numerator and Denominator by \[{z_1}\] in order to apply the value of \[\dfrac{{{z_2}}}{{{z_1}}} = \dfrac{{11}}{5}ik\].
\[\left| {\dfrac{{\left( {2{z_1} + 3{z_2}} \right)}}{{\left( {2{z_1} - 3{z_2}} \right)}}} \right|\]
After Dividing by \[{z_1}\], we will get ,
\[\left| {\dfrac{{2 + 3\dfrac{{{z_2}}}{{{z_1}}}}}{{2 - 3\dfrac{{{z_2}}}{{{z_1}}}}}} \right|\]
Substituting the values of \[\dfrac{{{z_2}}}{{{z_1}}}\] as \[\dfrac{{11}}{5}ik\]
\[ \Rightarrow \left| {\dfrac{{2 + 3\dfrac{{11}}{5}ik}}{{2 - 3\dfrac{{11}}{5}ik}}} \right|\]
\[ \Rightarrow \left| {\dfrac{{2 + \dfrac{{33}}{5}ik}}{{2 - \dfrac{{33}}{5}ik}}} \right|\]
According to the conjugate property of complex number ,
\[z = a + ib\]
\[\left| z \right| = \sqrt {{a^2} + {b^2}} \]
And similarly ,
\[z = a - ib\]
\[\left| z \right| = \sqrt {{a^2} + {b^2}} \]
Applying this concept in the mod,
\[ \Rightarrow \left| {\dfrac{{2 + \dfrac{{33}}{5}ik}}{{2 - \dfrac{{33}}{5}ik}}} \right|\]
\[ \Rightarrow \dfrac{{\sqrt {{2^2} + {{\left( {\dfrac{{33}}{5}k} \right)}^2}} }}{{\sqrt {{2^2} + {{\left( {\dfrac{{33}}{5}k} \right)}^2}} }}\]
\[ \Rightarrow 1\]
Hence the answer is (C) which is \[1\].

Note: Students must be very aware of the concept of the conjugate of complex numbers because without that concept it would be very tough to solve the question. Students should not get confused about the modulus of a complex number and its conjugate as both are the same.