JEE Main 2026 — Mathematics Algebra

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

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

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

$$\dfrac{iy+3}{iy-i} = \dfrac{3+iy}{i(y-1)} = \dfrac{-i(3+iy)}{y-1} = \dfrac{y - 3i}{y-1} = \dfrac{y}{y-1} - i\dfrac{3}{y-1}$$iy−iiy+3​=i(y−1)3+iy​=y−1−i(3+iy)​=y−1y−3i​=y−1y​−iy−13​

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

$$\dfrac{y}{y-1} = \dfrac{-3}{y-1} > 0$$y−1y​=y−1−3​>0

Since $y \neq 1$y=1, equating the numerators gives $y = -3$y=−3.

Checking for positivity: $\dfrac{-3}{-3-1} = \dfrac{3}{4} > 0$−3−1−3​=43​>0, which is valid.

Thus, $z = -3i$z=−3i.

The value of $|z|^2$∣z∣2 is $|-3i|^2 = 9$∣−3i∣2=9.

Answer: $9$9