Let Z_1 and Z_2 be complex numbers such that (3Z_1)/(4Z_2) is...

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|$.<br><br>Dividing the numerator and the denominator by$Z_2$, we get:<br><br>\left|\dfrac{\dfrac{Z_1}{Z_2} + 1}{\dfrac{Z_1}{Z_2} - 1}\right|$

Substituting $\dfrac{Z_1}{Z_2} = i y$:

\left|\dfrac{i y + 1}{i y - 1}\right| = \dfrac{|1 + i y|}{|-1 + i y|}$<br><br>Since$|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: $1$