If $\dfrac{z - \alpha}{z + \alpha}$ ($\alpha \in \mathbb{R}$) is a purely imaginary number and $|z| = 2$, then a value of $\alpha$ is
Step-by-Step Solution
Key Concept: For $\dfrac{z-\alpha}{z+\alpha}$ with $\alpha\in\mathbb{R}$ to be purely imaginary, its real part must vanish, which forces $|z|^2=\alpha^2$.
**Step 1: Condition for purely imaginary**
A complex number $w$ is purely imaginary iff $\text{Re}(w)=0$.
**Step 2: Compute real part**
Let $z=x+iy$. Multiply numerator and denominator by the conjugate of the denominator: $\text{Re}\!\left(\dfrac{z-\alpha}{z+\alpha}\right) = \dfrac{x^2+y^2-\alpha^2}{|z+\alpha|^2} = \dfrac{|z|^2-\alpha^2}{|z+\alpha|^2}$.
**Step 3: Set real part to zero**
$|z|^2-\alpha^2=0 \Rightarrow \alpha^2=|z|^2=4 \Rightarrow \alpha=\pm 2$. From the options, $\alpha=2$.
Correct Answer: 4