If `prod_(p=1)^(r) e^(iptheta)=1`, where `prod` denotes the continued product and `i=sqrt(-1)`, the most general value of `theta` is (where, n is an integer)

Options

A. (a)`(2npi)/(r(r-1)),n in I`
B. (b)`(2npi)/(r(r+1)),n in I`
C. (c)`(4npi)/(r(r-1)),n in I`
D. (d)`(4npi)/(r(r+1)),n in I`

Answer

To solve the problem, we start with the equation given:

\[ \prod_{p=1}^{r} e^{ip\theta} = 1 \]

### Step 1: Rewrite the Product
We can rewrite the product as:

\[ e^{i\theta} \cdot e^{i2\theta} \cdot e^{i3\theta} \cdots e^{ir\theta} \]

This can be simplified using the property of exponents:

\[ e^{i(\theta + 2\theta + 3\theta + \cdots + r\theta)} = e^{i\theta(1 + 2 + 3 + \ldots + r)} \]

### Step 2: Sum of First r Natural Numbers
The sum of the first \( r \) natural numbers is given by the formula:

\[ 1 + 2 + 3 + \ldots + r = \frac{r(r + 1)}{2} \]

### Step 3: Substitute the Sum into the Exponent
Now substitute this sum back into the exponent:

\[ e^{i\theta \cdot \frac{r(r + 1)}{2}} = 1 \]

### Step 4: Set the Exponent Equal to 2πn
For the expression \( e^{ix} = 1 \), the exponent must be an integer multiple of \( 2\pi \):

\[ \theta \cdot \frac{r(r + 1)}{2} = 2\pi n \]

where \( n \) is an integer.

### Step 5: Solve for θ
Now, we can solve for \( \theta \):

\[ \theta = \frac{2 \cdot 2\pi n}{r(r + 1)} = \frac{4\pi n}{r(r + 1)} \]

### Final Answer
Thus, the most general value of \( \theta \) is:

\[ \theta = \frac{4\pi n}{r(r + 1)} \]

where \( n \) is an integer.

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