Find the `theta` such that `(3+2i sin theta)/(1-2 isin theta)` is
(a) real
(b) purely imaginary

Answer

To solve the problem, we need to find the angle \( \theta \) such that the complex number

\[ \frac{3 + 2i \sin \theta}{1 - 2i \sin \theta} \]

is (a) real and (b) purely imaginary.

### Part (a): When the complex number is real

1. **Identify the imaginary part**: For the complex number to be real, its imaginary part must be zero. The imaginary part of the given complex number can be expressed as:
\[ \text{Imaginary part} = \frac{8 \sin \theta}{1 + 4 \sin^2 \theta} \]
Set this equal to zero:
\[ \frac{8 \sin \theta}{1 + 4 \sin^2 \theta} = 0 \]

2. **Solve for \( \sin \theta \)**: The fraction is zero when the numerator is zero (the denominator cannot be zero):
\[ 8 \sin \theta = 0 \implies \sin \theta = 0 \]
The solutions for \( \sin \theta = 0 \) are:
\[ \theta = n\pi \quad (n \in \mathbb{Z}) \]

### Part (b): When the complex number is purely imaginary

1. **Identify the real part**: For the complex number to be purely imaginary, its real part must be zero. The real part of the given complex number can be expressed as:
\[ \text{Real part} = \frac{3 - 4 \sin^2 \theta}{1 + 4 \sin^2 \theta} \]
Set this equal to zero:
\[ \frac{3 - 4 \sin^2 \theta}{1 + 4 \sin^2 \theta} = 0 \]

2. **Solve for \( \sin^2 \theta \)**: The fraction is zero when the numerator is zero:
\[ 3 - 4 \sin^2 \theta = 0 \implies 4 \sin^2 \theta = 3 \implies \sin^2 \theta = \frac{3}{4} \]

3. **Find \( \sin \theta \)**: Taking the square root gives:
\[ \sin \theta = \pm \frac{\sqrt{3}}{2} \]
The angles corresponding to this sine value are:
\[ \theta = n\pi + (-1)^n \frac{\pi}{3} \quad (n \in \mathbb{Z}) \]

### Summary of Solutions:
- For part (a), \( \theta = n\pi \) where \( n \) is an integer.
- For part (b), \( \theta = n\pi + (-1)^n \frac{\pi}{3} \) where \( n \) is an integer.