Q154 / 270 · NDA 2026 (Phase 1) · Mathematics
Let $Z_1$ and $Z_2$ be complex numbers such that $\dfrac{3Z_1}{4Z_2}$ is purely imaginary. What is $\left|\dfrac{Z_1+Z_2}{Z_1-Z_2}\right|$ equal to ?
- A$2$
- B$\dfrac{3}{2}$
- C$\dfrac{5}{4}$
- Correct answerD$1$
Solution
Given $\dfrac{3Z_1}{4Z_2}$ is purely imaginary.
Let $\dfrac{3Z_1}{4Z_2} = i k$, where $k \in \mathbb{R}$.
$\Rightarrow \dfrac{Z_1}{Z_2} = \dfrac{4}{3} i k$
Let $\dfrac{4}{3} k = y$, so $\dfrac{Z_1}{Z_2} = i y$, where $y \in \mathbb{R}$.
We need to evaluate $\left|\dfrac{Z_1+Z_2}{Z_1-Z_2}\right|$.
Dividing the numerator and the denominator by $Z_2$, we get:
$\left|\dfrac{\dfrac{Z_1}{Z_2} + 1}{\dfrac{Z_1}{Z_2} - 1}\right|$
Substituting $\dfrac{Z_1}{Z_2} = i y$:
$\left|\dfrac{i y + 1}{i y - 1}\right| = \dfrac{|1 + i y|}{|-1 + i y|}$
Since $|1 + i y| = \sqrt{1^2 + y^2}$ and $|-1 + i y| = \sqrt{(-1)^2 + y^2} = \sqrt{1 + y^2}$, we have:
$\dfrac{\sqrt{1 + y^2}}{\sqrt{1 + y^2}} = 1$
Answer: $1$
Let $\dfrac{3Z_1}{4Z_2} = i k$, where $k \in \mathbb{R}$.
$\Rightarrow \dfrac{Z_1}{Z_2} = \dfrac{4}{3} i k$
Let $\dfrac{4}{3} k = y$, so $\dfrac{Z_1}{Z_2} = i y$, where $y \in \mathbb{R}$.
We need to evaluate $\left|\dfrac{Z_1+Z_2}{Z_1-Z_2}\right|$.
Dividing the numerator and the denominator by $Z_2$, we get:
$\left|\dfrac{\dfrac{Z_1}{Z_2} + 1}{\dfrac{Z_1}{Z_2} - 1}\right|$
Substituting $\dfrac{Z_1}{Z_2} = i y$:
$\left|\dfrac{i y + 1}{i y - 1}\right| = \dfrac{|1 + i y|}{|-1 + i y|}$
Since $|1 + i y| = \sqrt{1^2 + y^2}$ and $|-1 + i y| = \sqrt{(-1)^2 + y^2} = \sqrt{1 + y^2}$, we have:
$\dfrac{\sqrt{1 + y^2}}{\sqrt{1 + y^2}} = 1$
Answer: $1$
Hence, the correct answer is option: 'D'.