If `(pi)/(2) lt theta lt (3 pi)/(2)` , the modulus and argument form of `(1 + cos 2 theta) + i sin 2 theta ` is

Options

A. `- 2 cos theta [ cos ( pi + theta) + isin ( pi + theta) ]`
B. `2 cos theta [ cos theta +isin theta]`
C. ` 2 cos theta [ cos theta + i sin (- theta) ]`
D. `- 2 cos theta {cos (pi - theta) + isin ( pi - theta)}`

Answer

To find the modulus and argument form of the complex number \( z = (1 + \cos 2\theta) + i \sin 2\theta \) where \( \frac{\pi}{2} < \theta < \frac{3\pi}{2} \), we will follow these steps:

### Step 1: Identify the real and imaginary parts
The complex number can be expressed as:
\[ z = (1 + \cos 2\theta) + i \sin 2\theta \]
Here, the real part \( x = 1 + \cos 2\theta \) and the imaginary part \( y = \sin 2\theta \).

### Step 2: Calculate the modulus of \( z \)
The modulus of a complex number \( z = x + iy \) is given by:
\[ |z| = \sqrt{x^2 + y^2} \]
Substituting the values of \( x \) and \( y \):
\[ |z| = \sqrt{(1 + \cos 2\theta)^2 + (\sin 2\theta)^2} \]

### Step 3: Expand the expression for modulus
Now, we will expand \( (1 + \cos 2\theta)^2 \):
\[ (1 + \cos 2\theta)^2 = 1 + 2\cos 2\theta + \cos^2 2\theta \]
Using the identity \( \sin^2 2\theta + \cos^2 2\theta = 1 \), we can substitute \( \sin^2 2\theta \) for \( 1 - \cos^2 2\theta \):
\[ |z| = \sqrt{(1 + 2\cos 2\theta + \cos^2 2\theta) + \sin^2 2\theta} \]
\[ = \sqrt{1 + 2\cos 2\theta + 1} = \sqrt{2 + 2\cos 2\theta} \]
\[ = \sqrt{2(1 + \cos 2\theta)} = \sqrt{2} \sqrt{1 + \cos 2\theta} \]

### Step 4: Simplify using the double angle identity
Using the identity \( 1 + \cos 2\theta = 2\cos^2 \theta \):
\[ |z| = \sqrt{2} \sqrt{2\cos^2 \theta} = 2|\cos \theta| \]
Since \( \theta \) is in the interval \( \left(\frac{\pi}{2}, \frac{3\pi}{2}\right) \), \( \cos \theta < 0 \), thus:
\[ |z| = -2\cos \theta \]

### Step 5: Calculate the argument of \( z \)
The argument of \( z \) is given by:
\[ \arg(z) = \tan^{-1}\left(\frac{y}{x}\right) = \tan^{-1}\left(\frac{\sin 2\theta}{1 + \cos 2\theta}\right) \]
Using the double angle identity, \( \sin 2\theta = 2\sin \theta \cos \theta \) and \( 1 + \cos 2\theta = 2\cos^2 \theta \):
\[ \arg(z) = \tan^{-1}\left(\frac{2\sin \theta \cos \theta}{2\cos^2 \theta}\right) = \tan^{-1}\left(\frac{\sin \theta}{\cos \theta}\right) = \tan^{-1}(\tan \theta) = \theta \]

### Step 6: Adjust the argument based on the quadrant
Since \( \theta \) is in the interval \( \left(\frac{\pi}{2}, \frac{3\pi}{2}\right) \), the angle will be in the second or third quadrant. Therefore, we need to add \( \pi \) to the argument:
\[ \arg(z) = \theta + \pi \]

### Final Result
Combining the modulus and argument, we can express \( z \) in modulus-argument form:
\[ z = -2\cos \theta \cdot e^{i(\theta + \pi)} \]