What is angle \theta such that z is purely real, 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 only

Correct Answer: C. n\pi

Explanation

Multiplying the numerator and denominator by the conjugate 1+i\sin\theta, we get z = \frac{(1+i\sin\theta)^2}{1+\sin^2\theta} = \frac{1-\sin^2\theta + 2i\sin\theta}{1+\sin^2\theta}. For z to be purely real, the imaginary part must be zero. So, \frac{2\sin\theta}{1+\sin^2\theta} = 0 \implies \sin\theta = 0. This gives \theta = n\pi.

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