JEE Main Mathematics Question (2026) — Solution

Step-by-step 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 R . We are given ( z+3 z-i ) = 4 . Substituting z = iy, we get: iy+3 iy-i = 3+iy i(y-1) = -i(3+iy) y-1 = y - 3i y-1 = y y-1 - i 3 y-1 For a complex number X + iY to have an argument of 4 , its real and imaginary parts must be equal and strictly positive. Therefore: y y-1 = -3 y-1 > 0 Since y 1, equating the numerators gives y = -3. Checking for positivity: -3 -3-1 = 3 4 > 0, which is valid. Thus, z = -3i. The value of |z|^2 is |-3i|^2 = 9. Answer: 9