Question

If $\dfrac{5{{z}_{2}}}{7{{z}_{1}}}$ is purely imaginary, then $\left| \dfrac{2{{z}_{1}}+3{{z}_{2}}}{2{{z}_{1}}-3{{z}_{2}}} \right|$ to
(a) \[\dfrac{5}{7}\]
(b) \[\dfrac{7}{9}\]
(c) \[\dfrac{25}{49}\]
(d) 1

Answer 1

Hint: To find the value of $\left| \dfrac{2{{z}_{1}}+3{{z}_{2}}}{2{{z}_{1}}-3{{z}_{2}}} \right|$, we have to convert it in the form of \[\dfrac{{{z}_{2}}}{{{z}_{1}}}\] using appropriate operations, because we know the values of \[\dfrac{{{z}_{2}}}{{{z}_{1}}}\]. Also, we have to make use of the formula |z|= \[\sqrt{{{a}^{2}}+{{b}^{2}}}\] where |z| is called modulus of \[z=a+ib\]

Complete step by step answer:
The question demands that, we have to find the value of the term $\left| \dfrac{2{{z}_{1}}+3{{z}_{2}}}{2{{z}_{1}}-3{{z}_{2}}} \right|$. Let the value of this term be ‘y’. Therefore, we will get,
\[y=\left| \dfrac{2{{z}_{1}}+3{{z}_{2}}}{2{{z}_{1}}-3{{z}_{2}}} \right|.............(i)\]
We are also given in question that $\dfrac{5{{z}_{2}}}{7{{z}_{1}}}$ is purely imaginary. This means we can say that we can represent $\dfrac{5{{z}_{2}}}{7{{z}_{1}}}$ solely in terms of I (iota). Thus,
$\dfrac{5{{z}_{2}}}{7{{z}_{1}}}$= \[ki\]
Where, k is any real number. We can also write the above equation as:
\[\dfrac{{{z}_{2}}}{{{z}_{1}}}=\dfrac{7ki}{5}...............(ii)\]
Now we come back to equation (i). Now we will divide both the numerator and denominator by ‘z’. After doing this we get: -
\[y=\left| \begin{align}
  & \dfrac{2{{z}_{1}}+3{{z}_{x}}}{{{z}_{1}}} \\
 & \overline{\,\,\dfrac{2{{z}_{1}}+3{{z}_{z}}}{{{z}_{1}}}} \\
\end{align} \right|\]
\[\Rightarrow y=\left| \dfrac{2+3\left( \dfrac{{{z}_{2}}}{{{z}_{1}}} \right)}{2-3\left( \dfrac{{{z}_{2}}}{{{z}_{1}}} \right)} \right|..................(iii)\]
Now, we will substitute value of \[\left( \dfrac{{{z}_{2}}}{{{z}_{1}}} \right)\] from equation (ii) into equation (iii). After doing this we will get:
$\Rightarrow y=\left| \dfrac{2+3\left( \dfrac{7{{k}_{1}}}{5} \right)}{2-3\left( \dfrac{7{{k}_{1}}}{5} \right)} \right|$
On simplifying we will get: -
$\Rightarrow y=\left| \dfrac{2+\dfrac{21ki}{5}}{2-\dfrac{21ki}{5}} \right|$
On taking Lcm and cancelling 5 from both numerator and denominator, we get: -
$\Rightarrow y=\left| \dfrac{10+21ki}{10-21ki} \right|.............(iv)$
Here, we are going to use a property of modulus which is shown below:
$\dfrac{|{{z}_{1}}|}{|{{z}_{2}}|}=\dfrac{|{{z}_{1}}|}{|{{z}_{2}}|}$
In our case ${{z}_{1}}=10+21ki$ and ${{z}_{2}}=10-21ki$
After using this identity, we will get:
$\Rightarrow y=\dfrac{|10+21ki|}{|10-21ki|}....................(v)$
In the above equation, we have to find the modulus of two terms in numerator and denominator respectively. If a complex number is z=a+ib then its modulus is given by the formula:
$|z|=\sqrt{{{a}^{2}}+{{b}^{2}}}$
Therefore, using above formula, we get: -
\[\begin{align}
  & \Rightarrow y=\dfrac{\sqrt{{{\left( 10 \right)}^{2}}+{{\left( 21k \right)}^{2}}}}{\sqrt{{{\left( 10 \right)}^{2}}+{{\left( -21k \right)}^{2}}}} \\
 & \Rightarrow y=\dfrac{\sqrt{100+441{{k}^{2}}}}{\sqrt{100+441{{k}^{2}}}} \\
 & \Rightarrow y=1 \\
 & \Rightarrow \left| \dfrac{2{{z}_{1}}+3{{z}_{2}}}{2{{z}_{1}}-3{{z}_{2}}} \right| \\
\end{align}\]

So, the correct answer is “Option d”.

Note: We cannot use the modulus formula directly in the starting as shown below: -
$\left| \dfrac{2{{z}_{1}}+3{{z}_{2}}}{2{{z}_{1}}-3{{z}_{2}}} \right|=\dfrac{\sqrt{{{\left( z \right)}^{2}}+{{\left( 3 \right)}^{2}}}}{\sqrt{{{\left( a \right)}^{2}}+{{\left( -3 \right)}^{2}}}}=\dfrac{\sqrt{13}}{\sqrt{13}}=1$
In this case, the answer is the same but the method is wrong because ${{z}_{1}}$ and ${{z}_{2}}$ are both complex numbers. We will have to convert the numerator and denominator into a single complex number of the form a+ib then only we can apply the modulus formula.