JEE Main 2026 — Complex Number Question with Solution

Given $|z+2|=|z-2|$, the point $z$ lies on the perpendicular bisector of the line segment joining $(-2,0)$ and $(2,0)$. This means $z$ lies on the imaginary axis.

Let $z=iy$, where $y\in \mathbb{R}$.

We are given $\arg \left(\frac{z+3}{z-i}\right)=\frac{\pi }{4}$. Substituting $z=iy$, we get:

$\frac{iy+3}{iy-i}=\frac{3+iy}{i(y-1)}=\frac{-i(3+iy)}{y-1}=\frac{y-3i}{y-1}=\frac{y}{y-1}-i\frac{3}{y-1}$

For a complex number $X+iY$ to have an argument of $\frac{\pi }{4}$, its real and imaginary parts must be equal and strictly positive. Therefore:

$\frac{y}{y-1}=\frac{-3}{y-1}>0$

Since $y\ne 1$, equating the numerators gives $y=-3$.

Checking for positivity: $\frac{-3}{-3-1}=\frac{3}{4}>0$, which is valid.

Thus, $z=-3i$.

The value of $|z{|}^2$ is $|-3i{|}^2=9$.

Answer: $9$