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

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: -

& \dfrac{2{{z}_{1}}+3{{z}_{x}}}{{{z}_{1}}} \\\ & \overline{\,\,\dfrac{2{{z}_{1}}+3{{z}_{z}}}{{{z}_{1}}}} \\\ \end{aligned} \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{aligned} & \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{aligned}$$ **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.