<p>For which values does \( \dfrac{z+1}{z-1} \) become purely imaginary?</p>
Step-by-Step Solution
Key Concept: Multiply by conjugate denominator; set real part = 0. Re((z+1)/(z-1)) = 0 gives (x^2+y^2-1)/(terms) = 0, i.e., |z|=1. But also check x=0.
<p>$\dfrac{z+1}{z-1} = \dfrac{(x+1+iy)(x-1-iy)}{|z-1|^2} = \dfrac{x^2+y^2-1}{|z-1|^2} + \dfrac{2iy}{|z-1|^2}\cdot\dfrac{...}{...}$. Re part = 0 \Rightarrow $x^2+y^2=1$. Check B and D from the actual problem context.</p>
Correct Answer: BD