JEE Main 2024 · MathsMultiple choiceSingle correctMediumMulti-step

JEE Main 9 April 2024, Shift 2, Maths Q2

Question 2 of 90 in this shift, Maths question 2 of 30, Section A.

Let

z

be a complex number such that the real part of

\frac{z-2i}{z+2i}

is zero. Then, the maximum value of

|z-(6+8i)|

is equal to

(1) $8$
(2) $10$
(3) $12$ Official answer
(4) $\infty$

Official answer

Option 3

NTA final key.

Chapter: Complex Numbers and Quadratic Equations
Topic: Complex Numbers - Polar Form
Idea tested: Polar Triangle Inequality Extrema

Same idea in other shifts

Asked 4× in all

2 Apr 2026, Shift 2 · Q2Let the circles $C_1: |z| = r$ and $C_2: |z - 3 - 4i| = 5, z \in \mathbb{C}_e$ be such that $C_2$ lies within $C_1$ . If $z_1$ moves on…MediumSingle correct
2 Apr 2025, Shift 1 · Q2Let $z$ be a complex number such that $|z|=1$ . If $\frac{2+k^2z}{k+\bar{z}}=kz$ , $k\in\mathbf{R}$ , then the maximum distance of $k+ik^2$ …HardSingle correct
28 Jul 2022, Shift 1 · Q15Let $S_1=\left\{z_1\in\mathbf{C}:|z_1-3|=\frac{1}{2}\right\}$ and $S_2=\{z_2\in\mathbf{C}:|z_2-|z_2+1||=|z_2+|z_2-1||\}$ . Then, for…HardSingle correct

Question text from the official JEE Main paper published by NTA; answer from the final answer key. Chapter, topic and idea tags, difficulty and skill are Lumi’s. Lumi analyses 42 of the 174 JEE Main shifts held from 2017 to 2026.