`(5 + i sin theta)/(5-3i sin theta)` is a real number when

Options

A. `theta = pi//4`
B. `theta = - pi`
C. `theta = - pi//2`
D. `theta = pi//2`

Answer

To determine when the expression \(\frac{5 + i \sin \theta}{5 - 3i \sin \theta}\) is a real number, we will follow these steps:

### Step 1: Understand the condition for a complex number to be real
A complex number is real if its imaginary part is zero. Therefore, we need to find conditions under which the imaginary part of the given expression is equal to zero.

### Step 2: Write the expression
We have:
\[ z = \frac{5 + i \sin \theta}{5 - 3i \sin \theta} \]

### Step 3: Rationalize the denominator
To eliminate the imaginary part from the denominator, we multiply the numerator and denominator by the conjugate of the denominator:
\[ z = \frac{(5 + i \sin \theta)(5 + 3i \sin \theta)}{(5 - 3i \sin \theta)(5 + 3i \sin \theta)} \]

### Step 4: Calculate the denominator
The denominator becomes:
\[ (5 - 3i \sin \theta)(5 + 3i \sin \theta) = 5^2 - (3i \sin \theta)^2 = 25 - 9 \sin^2 \theta \]

### Step 5: Calculate the numerator
The numerator expands to:
\[ (5 + i \sin \theta)(5 + 3i \sin \theta) = 25 + 15i \sin \theta + 5i \sin \theta - 3 \sin^2 \theta = 25 - 3 \sin^2 \theta + 20i \sin \theta \]

### Step 6: Combine the results
Thus, we can write:
\[ z = \frac{(25 - 3 \sin^2 \theta) + 20i \sin \theta}{25 - 9 \sin^2 \theta} \]

### Step 7: Identify the imaginary part
The imaginary part of \(z\) is:
\[ \text{Imaginary part} = \frac{20 \sin \theta}{25 - 9 \sin^2 \theta} \]

### Step 8: Set the imaginary part to zero
For \(z\) to be a real number, we set the imaginary part equal to zero:
\[ 20 \sin \theta = 0 \]

### Step 9: Solve for \(\theta\)
This implies:
\[ \sin \theta = 0 \]
The solutions for \(\sin \theta = 0\) are:
\[ \theta = n\pi \quad \text{where } n \in \mathbb{Z} \]

### Final Answer
The expression \(\frac{5 + i \sin \theta}{5 - 3i \sin \theta}\) is a real number when:
\[ \theta = n\pi \quad (n \text{ is any integer}) \]

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