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JEE Advanced 2024 Paper 1 · Mathematics · Complex Numbers · hard

Match the entries of List-I with those of List-II. List-I P. Number of complex numbers z satisfying z^3=bar z Q. Minimum value of |z-3-4i| over all z…

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Match the entries of List-I with those of List-II.

List-I

P. Number of complex numbers zz satisfying z3=zˉz^3=\bar z

Q. Minimum value of ∣z−3−4i∣|z-3-4i| over all zz with ∣z∣=1|z|=1

R. Maximum value of ∣z∣|z| over all zz with ∣z+1z∣=2\left|z+\dfrac1z\right|=2

S. Area of the triangle whose vertices are the roots of z3=8z^3=8

List-II

(1) 44

(2) 333\sqrt3

(3) 55

(4) 1+21+\sqrt2

(5) 22
  1. A. P→5, Q→1, R→4, S→2
  2. B. P→3, Q→1, R→2, S→4
  3. C. P→3, Q→5, R→4, S→2
  4. D. P→3, Q→1, R→4, S→2
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Answer: (D) P→3, Q→1, R→4, S→2

P: taking moduli, ∣z∣3=∣z∣|z|^3=|z| gives ∣z∣=0|z|=0 or 11. For ∣z∣=1|z|=1, z3=zˉ=1/zz^3=\bar z=1/z means z4=1z^4=1, giving 44 solutions; with z=0z=0 the total is 55, i.e. (3). Q: the point 3+4i3+4i is at distance 55 from the origin, so the least distance to the unit circle is 5−1=45-1=4, i.e. (1). R: for r=∣z∣r=|z| we have r−1r≤∣z+1z∣=2r-\tfrac1r\le\left|z+\tfrac1z\right|=2, so r2−2r−1≤0r^2-2r-1\le0 and r≤1+2r\le1+\sqrt2; at r=1+2r=1+\sqrt2 we have r2+r−2=6r^2+r^{-2}=6 and the condition r2+r−2+2cos⁡2θ=4r^2+r^{-2}+2\cos2\theta=4 is met with cos⁡2θ=−1\cos2\theta=-1, so the bound is attained: (4). S: the roots 2,2ω,2ω22,2\omega,2\omega^2 form an equilateral triangle of circumradius 22 and side 232\sqrt3, with area 34⋅12=33\tfrac{\sqrt3}{4}\cdot12=3\sqrt3, i.e. (2). So P→3, Q→1, R→4, S→2.
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