The general value of `theta` which satisfies the equation `(costheta+isintheta)(cos3theta+isin3theta) (cos5theta+isin5theta)…………((cos2n-1)theta+isin(2n-1)theta)` = 1 is

Options

A. `(rpi)/n^(2)`
B. `((r-1)pi)/(n^(2))`
C. `((2r+1)pi)/(n^(3))`
D. `(2rpi)/n^(2)`

Answer

To solve the equation

\[ (\cos \theta + i \sin \theta)(\cos 3\theta + i \sin 3\theta)(\cos 5\theta + i \sin 5\theta) \ldots (\cos (2n-1)\theta + i \sin (2n-1)\theta) = 1, \]

we can follow these steps:

### Step 1: Rewrite the terms using Euler's formula
Using Euler's formula, we can rewrite each term as:
\[ \cos k\theta + i \sin k\theta = e^{ik\theta}. \]
Thus, the left-hand side becomes:
\[ e^{i\theta} \cdot e^{i3\theta} \cdot e^{i5\theta} \cdots e^{i(2n-1)\theta}. \]

### Step 2: Combine the exponents
The product of the exponentials can be combined:
\[ e^{i(\theta + 3\theta + 5\theta + \ldots + (2n-1)\theta)} = e^{i\left(\sum_{k=0}^{n-1} (2k+1)\theta\right)}. \]

### Step 3: Calculate the sum of the series
The sum of the series \(1 + 3 + 5 + \ldots + (2n-1)\) is known to be \(n^2\). Therefore, we can write:
\[ \sum_{k=0}^{n-1} (2k+1) = n^2. \]
Thus, we have:
\[ e^{i(n^2 \theta)}. \]

### Step 4: Set the equation equal to 1
We now set the expression equal to 1:
\[ e^{i(n^2 \theta)} = 1. \]

### Step 5: Solve for \(\theta\)
The equation \(e^{i(n^2 \theta)} = 1\) implies that:
\[ n^2 \theta = 2k\pi, \]
where \(k\) is any integer.

### Step 6: Isolate \(\theta\)
Dividing both sides by \(n^2\), we find:
\[ \theta = \frac{2k\pi}{n^2}. \]

### Conclusion
Thus, the general value of \(\theta\) that satisfies the original equation is:
\[ \theta = \frac{2k\pi}{n^2}, \quad k \in \mathbb{Z}. \]