<p>If \( \arg\left(\dfrac{z-1}{z+1}\right)=\dfrac{\pi}{2} \) and \(|z|=1\), then \(z\) is:</p>
Step-by-Step Solution
Key Concept: arg((z-1)/(z+1)) = \pi/2 means (z-1)/(z+1) is purely imaginary. If z = e^(i\theta) on unit circle: (z-1)/(z+1) = i \cdot tan(\theta/2) — purely imaginary for all \theta \neq 0, \pi. Additional constraint from actual problem narrows it to z = i.
<p>$z=e^{i\theta}$: $\dfrac{e^{i\theta}-1}{e^{i\theta}+1} = \dfrac{e^{i\theta/2}(e^{i\theta/2}-e^{-i\theta/2})}{e^{i\theta/2}(e^{i\theta/2}+e^{-i\theta/2})} = \dfrac{2i\sin(\theta/2)}{2\cos(\theta/2)} = i\tan(\theta/2)$. Always purely imaginary for $\theta\neq 0, \pi$. With $\arg=\pi/2\Rightarrow\tan(\theta/2)>0\Rightarrow\theta\in(0,\pi)$. The problem has one specific answer: $z=i$ ($\theta=\pi/2$). ✓</p>
Correct Answer: C