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**Hint:**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 ${\text{i}}$, which is defined by its property ${{\text{i}}^{\text{2}}}{\text{ = - 1}}$.

**Formula used:**

${{\text{i}}^0}{\text{ = 1}}$

${{\text{i}}^{\text{1}}}{\text{ = i}}$

${{\text{i}}^{\text{2}}}{\text{ = - 1}}$

Absolute value square root theorem:

For any real number $x$, $\sqrt {{x^2}} = \left| x \right|$

**Complete step-by-step answer:**

Here it is a given that the expression is $\dfrac{{{\text{2}}{{\text{z}}_{\text{1}}}{\text{ + 3}}{{\text{z}}_{\text{2}}}}}{{{\text{2}}{{\text{z}}_{\text{1}}}{\text{ - 3}}{{\text{z}}_{\text{2}}}}}$. 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 ${\text{bi}}$ can be added to a real number ${\text{a}}$ to a complex number of the form ${\text{a + bi}}$.

Let \[\dfrac{{{\text{5}}{{\text{z}}_{\text{1}}}}}{{{\text{7}}{{\text{z}}_{\text{2}}}}} = {\text{iy}}\]

Here ${\text{iy}}$ is an imaginary number.

\[ \Rightarrow \dfrac{{{{\text{z}}_{\text{1}}}}}{{{{\text{z}}_{\text{2}}}}} = \dfrac{{{\text{7y}}}}{{\text{5}}}{\text{i}}\]

Consider the given term,

Here divide the given expression by ${{\text{z}}_{\text{2}}}$ we get,

$\left| {\dfrac{{{\text{2}}{{\text{z}}_{\text{1}}}{\text{ + 3}}{{\text{z}}_{\text{2}}}}}{{{\text{2}}{{\text{z}}_{\text{1}}}{\text{ - 3}}{{\text{z}}_{\text{2}}}}}} \right| = \left| {\dfrac{{{\text{2}}\dfrac{{{{\text{z}}_{\text{1}}}}}{{{{\text{z}}_{\text{2}}}}}{\text{ + 3}}}}{{{\text{2}}\dfrac{{{{\text{z}}_{\text{1}}}}}{{{{\text{z}}_{\text{2}}}}}{\text{ - 3}}}}} \right|$

Substitute the value for \[\dfrac{{{{\text{z}}_{\text{1}}}}}{{{{\text{z}}_{\text{2}}}}} = \dfrac{{{\text{7y}}}}{{\text{5}}}{\text{i}}\]

$ \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,

$ \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,

$\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,

\[ \Rightarrow {{\text{i}}^{\text{2}}}{\text{ = i}} \times {\text{i}}\]

Substituting \[{\text{i}} = \sqrt { - 1} \] we get,

\[ \Rightarrow \sqrt { - 1} \times \sqrt { - 1} {\text{ = }}{\left( {\sqrt { - 1} } \right)^2}\]

Simplifying the terms,

\[ \Rightarrow {\left( {\sqrt { - 1} } \right)^2} = - 1\]

Hence we get ${{\text{i}}^{\text{2}}} = - 1$ and ${14^2} = 14 \times 14 = 196$, ${5^2} = 5 \times 5 = 25$, ${3^2} = 3 \times 3 = 9$using these values we get,

$ \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,

$ \Rightarrow \sqrt 1 $

Hence we get,

$ \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 ($1$) 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 $n$, written $\left| n \right|$, can be described geometrically as the distance of $n$ from $0$ on the number line.

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