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AIEEE 2009

JEE Main 2009 Mathematics question on Complex Numbers and Quadratic Equations

Let z=cos⁡θ+isin⁡θz = \cos\theta + i\sin\theta. Then, the value of ∑m=115Im(z2m−1)\sum_{m=1}^{15} \text{Im}(z^{2m-1}) at θ=2°\theta = 2° is
  1. A.
    1sin⁡2°\dfrac{1}{\sin 2°}
  2. B.
    13sin⁡2°\dfrac{1}{3\sin 2°}
  3. C.
    12sin⁡2°\dfrac{1}{2\sin 2°}
  4. D.
    14sin⁡2°\dfrac{1}{4\sin 2°}
Show answer and solution

Correct option: D

Solution

Im(z^(2m-1)) = sin((2m-1)θ), so the sum is sin θ + sin 3θ + sin 5θ + ... + sin 29θ (15 terms of odd multiples).

Using the standard identity, the sum of sin((2m-1)θ) for m=1 to n equals sin²(nθ)/sin θ.

With n=15 and θ=2°: sin²(30°)/sin 2° = (1/2)²/sin 2° = (1/4)/sin 2° = 1/(4 sin 2°).

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