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 \[\sqrt{{{a}^{2}}+{{b}^{2}}}\]