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← JEE Advanced 2025 · Paper 2
MathematicsNumerical+4Question 13 of 48

JEE Advanced 2025 · Paper 2, Mathematics question 13

For a non-zero complex number zz, let arg⁡(z)\arg(z) denote the principal argument of zz, with −π<arg⁡(z)≤π-\pi < \arg(z) \le \pi. Let ω\omega be the cube root of unity for which 0<arg⁡(ω)<π0 < \arg(\omega) < \pi. Let

α=arg⁡(∑n=12025(−ω)n).\alpha = \arg\left(\sum_{n=1}^{2025}(-\omega)^n\right).

Then the value of 3απ\dfrac{3\alpha}{\pi} is _____.

Answer and worked solution

Official answer: -2

Sixth root of unity. −ω=eiπe2πi/3=e−iπ/3-\omega=e^{i\pi}e^{2\pi i/3}=e^{-i\pi/3}, so its powers repeat every 6 and each full block sums to 0.

Leftover terms. 2025=6⋅337+32025=6\cdot337+3, so the sum equals the first three terms:
−ω+ω2−1.-\omega+\omega^2-1.
Using 1+ω=−ω21+\omega=-\omega^2, this is 2ω22\omega^2.

Argument. arg⁡(2ω2)=−2π3\arg(2\omega^2)=-\frac{2\pi}3, so 3απ=−2\frac{3\alpha}\pi=-2.

Topics:Roots of unityModulus and argument
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