JEE Advanced 2025 Paper 2, Mathematics Section 3 Q5: De Moivre's Theorem And Its Applications

JEE Advanced2025Paper 2Mathematics Section 3

For a non-zero complex number $z$ , let $ar g (z)$ denote the principal argument of $z$ , with $- π < ar g (z) \leq π$ . Let $ω$ be the cube root of unity for which $0 < ar g (ω) < π$ . Let

$α = ar g (n = 1 \sum 2025 (- ω)^{n}) .$

Then the value of $\frac{3 α}{π}$ is ________.

Your answer

Solution

Step 1: Geometric series.

$n = 1 \sum 2025 (- ω)^{n} = \frac{- ω ( ( - ω ) ^{2025} - 1 )}{- ω - 1}$ .

Step 2: Simplify $(- ω)^{2025}$ .

$(- ω)^{2025} = - ω^{2025} = - (ω^{3})^{675} = - 1$ . Hence the numerator equals $- ω (- 1 - 1) = 2 ω$ .

Step 3: Use $1 + ω + ω^{2} = 0$ .

The denominator becomes $- (ω + 1) = ω^{2}$ . So the sum simplifies to $\frac{2 ω}{ω ^{2}} = \frac{2}{ω} = 2 ω^{2}$ (since $ω^{3} = 1$ ).

Step 4: Take the argument.

With $ω = e^{i 2 π /3}$ , we have $ω^{2} = e^{i 4 π /3} = e^{- i 2 π /3}$ . Therefore $ar g (2 ω^{2}) = - \frac{2 π}{3}$ .

Step 5: Final value.

$\frac{3 α}{π} = \frac{3 \cdot ( - 2 π /3 )}{π} = - 2$ .

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JEE Advanced 2025 Paper 2, Mathematics Section 3 Q5: De Moivre's Theorem And Its Applications

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