Complex Numbers
PYP_JEE_ADV_2025_P2
Grade None
Question:
For a non-zero complex number $z$, let $\text{arg}(z)$ denote the principal argument of $z$, with $-\pi < \text{arg}(z) \le \pi$. Let $\omega$ be the cube root of unity for which $0 < \text{arg}(\omega) < \pi$. Let
$$\alpha = \text{arg}\left(\sum_{n=1}^{2025} (-\omega)^n\right)$$
Then the value of $\dfrac{3\alpha}{\pi}$ is
Step-by-Step Solution
Key Concept: Summing geometric series of roots of unity, and correctly identifying the principal argument of a complex number in the third quadrant.
Given $0 < \text{arg}(\omega) < \pi \implies \omega = e^{i 2\pi/3} = -\dfrac{1}{2} + i\dfrac{\sqrt{3}}{2}$.
Thus: $-\omega = e^{-i \pi/3} = \dfrac{1}{2} - i\dfrac{\sqrt{3}}{2}$.
Let $r = -\omega = e^{-i \pi/3}$. The sum is $S = \sum_{n=1}^{2025} r^n$.
Note $r^6 = 1$. The number of terms is $2025 = 6 \times 337 + 3$.
Since the sum of 6 consecutive terms of a 6th root GP is zero:
$$S = 337(0) + (r + r^2 + r^3) = r + r^2 + r^3$$
Substitute values:
$$S = e^{-i\pi/3} + e^{-i2\pi/3} + e^{-i\pi}$$
$$= \left(\dfrac{1}{2} - i\dfrac{\sqrt{3}}{2}\right) + \left(-\dfrac{1}{2} - i\dfrac{\sqrt{3}}{2}\right) - 1$$
$$= -1 - i\sqrt{3}$$
Since $-1 - i\sqrt{3}$ lies in the third quadrant:
$$\alpha = \text{arg}(-1 - i\sqrt{3}) = -\pi + \tan^{-1}\left(\dfrac{-\sqrt{3}}{-1}\right) = -\pi + \dfrac{\pi}{3} = -\dfrac{2\pi}{3}$$
Hence: $\dfrac{3\alpha}{\pi} = \dfrac{3}{\pi}\left(-\dfrac{2\pi}{3}\right) = -2$.
Correct Answer: