JEE Main 2024 · MathsMultiple choiceSingle correctMediumMulti-step

JEE Main 8 April 2024, Shift 1, Maths Q2

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

Let

z

be a complex number such that

|z+2|=1

and

\mathrm{Im}\left(\frac{z+1}{z+2}\right)=\frac{1}{5}

. Then the value of

\left|\mathrm{Re}\left(\overline{z+2}\right)\right|

(1) $\frac{2\sqrt{6}}{5}$ Official answer
(2) $\frac{24}{5}$
(3) $\frac{\sqrt{6}}{5}$
(4) $\frac{1+\sqrt{6}}{5}$

Official answer

Option 1

NTA final key.

Chapter: Complex Numbers and Quadratic Equations
Topic: Complex Numbers - Polar Form
Idea tested: Polar Conjugate and Inverse

Same topic in other shifts

All Complex Numbers - Polar Form questions

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4 Apr 2026, Shift 1 · Q3Let $z$ be a complex number such that $|z+2|=|z-2|$ and $\arg\left(\frac{z+3}{z-i}\right)=\frac{\pi}{4}$ . Then $|z|^2$ is equal to: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
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6 Apr 2024, Shift 2 · Q3If $z_1, z_2$ are two distinct complex number such that $\left|\frac{z_1-2z_2}{\frac{1}{2}-z_1\bar{z_2}}\right|=2$ , thenHardSingle correct
9 Apr 2024, Shift 2 · Q2Let $z$ 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 toMediumSingle 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.