What is angle \theta such that z is purely imaginary, where n is an integer?

Consider the following for the next three (03) items that follow : Let z=\frac{1+i~\sin~\theta}{1-i~\sin~\theta} where i=\sqrt{-1}

  1. A. \frac{n\pi}{2}
  2. B. \frac{(2n+1)\pi}{2} ✓
  3. C. n\pi
  4. D. 2n\pi

Correct Answer: B. \frac{(2n+1)\pi}{2}

Explanation

From the rationalized form z = \frac{1-\sin^2\theta + 2i\sin\theta}{1+\sin^2\theta}, for z to be purely imaginary, the real part must be zero. Therefore, 1-\sin^2\theta = 0 \implies \sin^2\theta = 1. This implies \sin\theta = \pm 1, which corresponds to \theta = \frac{(2n+1)\pi}{2} for any integer n.

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