Which of the following is true for `z=(3+2isintheta)(1-2sintheta)w h e r ei=sqrt(-1)` ? z is purely real for `theta=npi+-pi/3,n in Z` z is purely imaginary for `theta=npi+-pi/2,n in Z` z is purely real for `theta=npi,n in Z` none of these

Options

A. z is purely real for `theta=n pi pm pi//3, n in Z`
B. z is purely imaginary for `theta=npi pm pi//2, n in Z`
C. z is purely real for `theta=n pi, n in Z`
D. none of these

Explanation

To solve the problem, we need to analyze the expression for \( z \) given by:

\[ z = (3 + 2i \sin \theta)(1 - 2 \sin \theta) \]

where \( i = \sqrt{-1} \).

### Step 1: Expand the expression for \( z \)

We will first expand the expression:

\[ z = (3 + 2i \sin \theta)(1 - 2 \sin \theta) \]

Using the distributive property (FOIL method):

\[ z = 3(1) + 3(-2 \sin \theta) + 2i \sin \theta(1) + 2i \sin \theta(-2 \sin \theta) \]

This simplifies to:

\[ z = 3 - 6 \sin \theta + 2i \sin \theta - 4i \sin^2 \theta \]

### Step 2: Combine real and imaginary parts

Now we can separate the real and imaginary parts of \( z \):

\[ z = (3 - 6 \sin \theta) + i(2 \sin \theta - 4 \sin^2 \theta) \]

Thus, the real part \( \text{Re}(z) \) is:

\[ \text{Re}(z) = 3 - 6 \sin \theta \]

And the imaginary part \( \text{Im}(z) \) is:

\[ \text{Im}(z) = 2 \sin \theta - 4 \sin^2 \theta \]

### Step 3: Determine when \( z \) is purely real

For \( z \) to be purely real, the imaginary part must be zero:

\[ 2 \sin \theta - 4 \sin^2 \theta = 0 \]

Factoring out \( 2 \sin \theta \):

\[ 2 \sin \theta (1 - 2 \sin \theta) = 0 \]

This gives us two cases:

1. \( 2 \sin \theta = 0 \) which implies \( \sin \theta = 0 \) leading to \( \theta = n\pi \) where \( n \in \mathbb{Z} \).
2. \( 1 - 2 \sin \theta = 0 \) which implies \( \sin \theta = \frac{1}{2} \) leading to \( \theta = n\pi \pm \frac{\pi}{6} \).

### Step 4: Determine when \( z \) is purely imaginary

For \( z \) to be purely imaginary, the real part must be zero:

\[ 3 - 6 \sin \theta = 0 \]

Solving for \( \sin \theta \):

\[ 6 \sin \theta = 3 \implies \sin \theta = \frac{1}{2} \]

This gives:

\[ \theta = n\pi \pm \frac{\pi}{6} \]

### Step 5: Conclusion

From our analysis, we find:

- \( z \) is purely real for \( \theta = n\pi \) and \( \theta = n\pi \pm \frac{\pi}{6} \).
- \( z \) is purely imaginary for \( \theta = n\pi \pm \frac{\pi}{6} \).

Thus, the correct answer is:

**Option C**: \( z \) is purely real for \( \theta = n\pi, n \in \mathbb{Z} \).