Question

If $\dfrac{5{{z}_{2}}}{11{{z}_{1}}}$ is purely imaginary, then $\left| \dfrac{2{{z}_{1}}+3{{z}_{2}}}{2{{z}_{1}}-3{{z}_{2}}} \right|$ is equal to
1. $\dfrac{37}{33}$
2. $1$
3. $2$
4. $3$

Answer 1

Hint : To solve this we must know certain things like a complex number is said to be purely imaginary if the real part of it is non-existent i.e. $0$ similarly a complex number is said to be purely real if the imaginary part of it is $0$. We must know that i is used to represent imaginary numbers where ${{i}^{2}}=1$ and $z=a+ib$ where a stands for the real part and b stands for imaginary part of the equation. $|z|$ is equal to $\sqrt{{{a}^{2}}+{{b}^{2}}}$. By using these concepts we can find the answer of $\left| \dfrac{2{{z}_{1}}+3{{z}_{2}}}{2{{z}_{1}}-3{{z}_{2}}} \right|$

Complete step-by-step answer :
Now to solve this we will first start with noticing that the expression we are given is completely imaginary so x will not exist and be zero. Therefore we can write the equation given to us as
$\dfrac{5{{z}_{2}}}{11{{z}_{1}}}=iy$
Cross multiplying this equation we get the value of ratio of both complex number which we can use further to simplify the question
$\dfrac{{{z}_{2}}}{{{z}_{1}}}=\dfrac{11iy}{5}$
Now we need to find the modulus of $\dfrac{2{{z}_{1}}+3{{z}_{2}}}{2{{z}_{1}}-3{{z}_{2}}}$ therefore we can write this expression as
$\Rightarrow \dfrac{{{z}_{1}}(2+3\dfrac{{{z}_{2}}}{{{z}_{1}}})}{{{z}_{1}}(2-3\dfrac{{{z}_{2}}}{{{z}_{1}}})}$
We divided both the numerator and denominator with ${{z}_{1}}$to get this expression. Now we can cancel ${{z}_{1}}$from numerator and denominator
$\Rightarrow \dfrac{2+3\dfrac{{{z}_{2}}}{{{z}_{1}}}}{2-3\dfrac{{{z}_{2}}}{{{z}_{1}}}}$
Substituting the value of the ratio of complex numbers we get,
$\Rightarrow \dfrac{2+3\times \dfrac{11iy}{5}}{2-3\dfrac{11iy}{5}}$
Simplifying we can write this as
$\Rightarrow \dfrac{10+33iy}{10-33iy}$
Now since we need $\left| \dfrac{2{{z}_{1}}+3{{z}_{2}}}{2{{z}_{1}}-3{{z}_{2}}} \right|$ taking modulus of both complex numbers
$\Rightarrow \dfrac{|10+33iy|}{|10-33iy|}$
Using the formula for $|z|$ we get
$\Rightarrow \dfrac{\sqrt{100+1089}}{\sqrt{100+1089}}$
Hence we can say that $\left| \dfrac{2{{z}_{1}}+3{{z}_{2}}}{2{{z}_{1}}-3{{z}_{2}}} \right|=1$
So, the correct answer is “Option 2”.

Note: There is a common mistake made always that the modulus of $z=a+ib$ will be $\sqrt{{{a}^{2}}-{{b}^{2}}}$. Make sure you avoid this mistake from happening or else the solution will be wrong , The modulus no matter will be equal to $\sqrt{{{a}^{2}}+{{b}^{2}}}$