Question

Sol. [Key Concept: $|z + 2| = |z - 2|$ puts $z$ on the imaginary axis; use the argument condition to fix $z$.]

$${|z + 2| = |z - 2| }{\Rightarrow z = 0 + iy\ y \in R }{arg\left( \frac{3 + iy}{iy - i} \right) = \frac{\pi}{4} }{\Rightarrow arg\left( \frac{y}{y - 1} + \frac{3i}{1 - y} \right) = \frac{\pi}{4} }{\Rightarrow \frac{y}{y - 1} = \frac{3}{1 - y} > 0 }{\Rightarrow z = - 3i }{|z|^{2} = 9 }$$