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Question: Find the value of \(\dfrac{{{\text{2}}{{\text{z}}_{\text{1}}}{\text{ + 3}}{{\text{z}}_{\text{2}}}}}{...

Find the value of 2z1 + 3z22z1 - 3z2\dfrac{{{\text{2}}{{\text{z}}_{\text{1}}}{\text{ + 3}}{{\text{z}}_{\text{2}}}}}{{{\text{2}}{{\text{z}}_{\text{1}}}{\text{ - 3}}{{\text{z}}_{\text{2}}}}} where 5z17z2\dfrac{{{\text{5}}{{\text{z}}_{\text{1}}}}}{{{\text{7}}{{\text{z}}_{\text{2}}}}} is purely imaginary.
A) 57\dfrac{5}{7}
B) 75\dfrac{7}{5}
C) 11
D) 35\dfrac{3}{5}

Explanation

Solution

In this question we have found the value of the given term. For that we are going to solve the given expression by using an imaginary number. An imaginary number is a complex number that can be written as a real number multiplied by the imaginary unit i{\text{i}}, which is defined by its property i2 = - 1{{\text{i}}^{\text{2}}}{\text{ = - 1}}.

Formula used:
i0 = 1{{\text{i}}^0}{\text{ = 1}}
i1 = i{{\text{i}}^{\text{1}}}{\text{ = i}}
i2 = - 1{{\text{i}}^{\text{2}}}{\text{ = - 1}}
Absolute value square root theorem:
For any real number xx, x2=x\sqrt {{x^2}} = \left| x \right|

Complete step-by-step answer:
Here it is a given that the expression is 2z1 + 3z22z1 - 3z2\dfrac{{{\text{2}}{{\text{z}}_{\text{1}}}{\text{ + 3}}{{\text{z}}_{\text{2}}}}}{{{\text{2}}{{\text{z}}_{\text{1}}}{\text{ - 3}}{{\text{z}}_{\text{2}}}}}. We have to find the value of the given expression. They give a complex number to solve this express. So by using the number we are going to solve and find the value of the given expression.
An imaginary number bi{\text{bi}} can be added to a real number a{\text{a}} to a complex number of the form a + bi{\text{a + bi}}.
Let 5z17z2=iy\dfrac{{{\text{5}}{{\text{z}}_{\text{1}}}}}{{{\text{7}}{{\text{z}}_{\text{2}}}}} = {\text{iy}}
Here iy{\text{iy}} is an imaginary number.
z1z2=7y5i\Rightarrow \dfrac{{{{\text{z}}_{\text{1}}}}}{{{{\text{z}}_{\text{2}}}}} = \dfrac{{{\text{7y}}}}{{\text{5}}}{\text{i}}
Consider the given term,
Here divide the given expression by z2{{\text{z}}_{\text{2}}} we get,
2z1 + 3z22z1 - 3z2=2z1z2 + 32z1z2 - 3\left| {\dfrac{{{\text{2}}{{\text{z}}_{\text{1}}}{\text{ + 3}}{{\text{z}}_{\text{2}}}}}{{{\text{2}}{{\text{z}}_{\text{1}}}{\text{ - 3}}{{\text{z}}_{\text{2}}}}}} \right| = \left| {\dfrac{{{\text{2}}\dfrac{{{{\text{z}}_{\text{1}}}}}{{{{\text{z}}_{\text{2}}}}}{\text{ + 3}}}}{{{\text{2}}\dfrac{{{{\text{z}}_{\text{1}}}}}{{{{\text{z}}_{\text{2}}}}}{\text{ - 3}}}}} \right|
Substitute the value for z1z2=7y5i\dfrac{{{{\text{z}}_{\text{1}}}}}{{{{\text{z}}_{\text{2}}}}} = \dfrac{{{\text{7y}}}}{{\text{5}}}{\text{i}}
27yi5 + 327yi5 - 3\Rightarrow \left| {\dfrac{{{\text{2}}\dfrac{{{\text{7yi}}}}{{\text{5}}}{\text{ + 3}}}}{{{\text{2}}\dfrac{{{\text{7yi}}}}{{\text{5}}}{\text{ - 3}}}}} \right|
Here multiplying the terms inside elements,
14yi5 + 314yi5 - 3\Rightarrow \left| {\dfrac{{\dfrac{{{\text{14yi}}}}{{\text{5}}}{\text{ + 3}}}}{{\dfrac{{{\text{14yi}}}}{{\text{5}}}{\text{ - 3}}}}} \right|
By using the absolute value square root theorem mentioned in formula used, we get,
14yi5 + 314yi5 - 3=(14yi5)2+(3)2(14yi5)2+(3)2\left| {\dfrac{{\dfrac{{{\text{14yi}}}}{{\text{5}}}{\text{ + 3}}}}{{\dfrac{{{\text{14yi}}}}{{\text{5}}}{\text{ - 3}}}}} \right| = \sqrt {\dfrac{{{{\left( {\dfrac{{14yi}}{5}} \right)}^2} + {{\left( 3 \right)}^2}}}{{{{\left( {\dfrac{{14yi}}{5}} \right)}^2} + {{( - 3)}^2}}}}
Consider an imaginary number,
i2 = i×i\Rightarrow {{\text{i}}^{\text{2}}}{\text{ = i}} \times {\text{i}}
Substituting i=1{\text{i}} = \sqrt { - 1} we get,
1×1 = (1)2\Rightarrow \sqrt { - 1} \times \sqrt { - 1} {\text{ = }}{\left( {\sqrt { - 1} } \right)^2}
Simplifying the terms,
(1)2=1\Rightarrow {\left( {\sqrt { - 1} } \right)^2} = - 1
Hence we get i2=1{{\text{i}}^{\text{2}}} = - 1 and 142=14×14=196{14^2} = 14 \times 14 = 196, 52=5×5=25{5^2} = 5 \times 5 = 25, 32=3×3=9{3^2} = 3 \times 3 = 9using these values we get,
(196y225)+9(196y225)+9\Rightarrow \sqrt {\dfrac{{\left( { - \dfrac{{196{y^2}}}{{25}}} \right) + 9}}{{\left( { - \dfrac{{196{y^2}}}{{25}}} \right) + 9}}}
Cancelling the similar terms we get,
1\Rightarrow \sqrt 1
Hence we get,
2z1 + 3z22z1 - 3z2=1\Rightarrow \left| {\dfrac{{{\text{2}}{{\text{z}}_{\text{1}}}{\text{ + 3}}{{\text{z}}_{\text{2}}}}}{{{\text{2}}{{\text{z}}_{\text{1}}}{\text{ - 3}}{{\text{z}}_{\text{2}}}}}} \right| = 1

\therefore The option C (11) is a correct answer.

Note: Here we are doing so many calculations using complex numbers. So we have to be careful on that calculation. Important thing is that we use absolute value-square root theorem. We have to know about the absolute value.
Absolute value: The absolute value of a number nn, written n\left| n \right|, can be described geometrically as the distance of nn from 00 on the number line.