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JEE Advanced 2025 Paper 1 · Mathematics · Complex Numbers · medium

Let ω be a non-real cube root of unity. Match each entry of List-I with its value in List-II. List-I (P) The number of complex numbers z satisfying…

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Let ω\omega be a non-real cube root of unity. Match each entry of List-I with its value in List-II.

List-I
(P) The number of complex numbers zz satisfying z5=zˉz^{5}=\bar z
(Q) The number of real solutions of 2x=x22^{x}=x^{2}
(R) The value of (1−ω+ω2)(1+ω−ω2)(1-\omega+\omega^{2})(1+\omega-\omega^{2})
(S) The value of ∑k=05cos⁡kπ3\displaystyle\sum_{k=0}^{5}\cos\frac{k\pi}{3}

List-II
(1) 00 (2) 11 (3) 33 (4) 44 (5) 77
  1. A. P→5, Q→4, R→3, S→1
  2. B. P→3, Q→5, R→4, S→1
  3. C. P→5, Q→3, R→4, S→1
  4. D. P→5, Q→3, R→1, S→4
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Answer: (C) P→5, Q→3, R→4, S→1

(P) Taking moduli, ∣z∣5=∣z∣|z|^5=|z|, so z=0z=0 or ∣z∣=1|z|=1. For ∣z∣=1|z|=1, multiplying by zz gives z6=∣z∣2=1z^6=|z|^2=1, which has 66 roots; with z=0z=0 there are 77, so P→5. (Q) 2x=x22^x=x^2 holds at x=2x=2 and x=4x=4, and also for one negative x≈−0.767x\approx-0.767, because 2x−x22^x-x^2 changes sign between −1-1 and 00; for x>4x>4, 2x2^x exceeds x2x^2 forever, so there are exactly 33 solutions: Q→3. (R) Using 1+ω+ω2=01+\omega+\omega^2=0: 1−ω+ω2=−2ω1-\omega+\omega^2=-2\omega and 1+ω−ω2=−2ω21+\omega-\omega^2=-2\omega^2, so the product is 4ω3=44\omega^3=4: R→4. (S) The six values cos⁡kπ3\cos\frac{k\pi}3 are 1,12,−12,−1,−12,121,\frac12,-\frac12,-1,-\frac12,\frac12 and sum to 00: S→1. The unmatched item (2) is a distractor. A numerical check of each count and value agrees.
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