Question

If $${\displaystyle \frac{5z_2}{11z_1}}$$ is purely imaginary, then $$\left|{\displaystyle \frac{2z_1+3z_2}{2z_1-3z_2}}\right|$$ is equal to
1. $${\displaystyle \frac{37}{33}}$$
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 $$\left|{\displaystyle \frac{2z_1+3z_2}{2z_1-3z_2}}\right|$$

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
$${\displaystyle \frac{5z_2}{11z_1}}=iy$$
Cross multiplying this equation we get the value of ratio of both complex number which we can use further to simplify the question
$${\displaystyle \frac{z_2}{z_1}}={\displaystyle \frac{11iy}{5}}$$
Now we need to find the modulus of $${\displaystyle \frac{2z_1+3z_2}{2z_1-3z_2}}$$ therefore we can write this expression as
$$\Rightarrow {\displaystyle \frac{z_1(2+3{\displaystyle \frac{z_2}{z_1}})}{z_1(2-3{\displaystyle \frac{z_2}{z_1}})}}$$
We divided both the numerator and denominator with $$z_1$$to get this expression. Now we can cancel $$z_1$$from numerator and denominator
$$\Rightarrow {\displaystyle \frac{2+3{\displaystyle \frac{z_2}{z_1}}}{2-3{\displaystyle \frac{z_2}{z_1}}}}$$
Substituting the value of the ratio of complex numbers we get,
$$\Rightarrow {\displaystyle \frac{2+3\times {\displaystyle \frac{11iy}{5}}}{2-3{\displaystyle \frac{11iy}{5}}}}$$
Simplifying we can write this as
$$\Rightarrow {\displaystyle \frac{10+33iy}{10-33iy}}$$
Now since we need $$\left|{\displaystyle \frac{2z_1+3z_2}{2z_1-3z_2}}\right|$$ taking modulus of both complex numbers
$$\Rightarrow {\displaystyle \frac{|10+33iy|}{|10-33iy|}}$$
Using the formula for $$|z|$$ we get
$$\Rightarrow {\displaystyle \frac{\sqrt{100+1089}}{\sqrt{100+1089}}}$$
Hence we can say that $$\left|{\displaystyle \frac{2z_1+3z_2}{2z_1-3z_2}}\right|=1$$
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 $$\sqrt{a^2+b^2}$$