If `(3pi)/(2) gt alpha gt 2 pi`, find the modulus and argument of `(1 - cos 2 alpha) + i sin 2 alpha `.

Options

A. `-2cosalpha[cos(pi+alpha)+isin(pi+alpha)]`
B. `2cosalpha[cosalpha+isinalpha}`
C. `2cosalpha[cos(pi-alpha)+isin(pi-alpha)}`
D. `-2cosalpha[cos(pi-alpha)+isin(pi-alpha)}`

Answer

To find the modulus and argument of the complex number \( z = (1 - \cos 2\alpha) + i \sin 2\alpha \), given that \( \frac{3\pi}{2} > \alpha > 2\pi \), we can follow these steps:

### Step 1: Use Trigonometric Identities
We start by using the trigonometric identities for \( \cos 2\alpha \) and \( \sin 2\alpha \):
- \( \cos 2\alpha = 1 - 2\sin^2 \alpha \)
- \( \sin 2\alpha = 2\sin \alpha \cos \alpha \)

### Step 2: Substitute the Identities
Substituting these identities into the expression for \( z \):
\[ z = (1 - (1 - 2\sin^2 \alpha)) + i(2\sin \alpha \cos \alpha) \]
This simplifies to:
\[ z = 2\sin^2 \alpha + i(2\sin \alpha \cos \alpha) \]

### Step 3: Factor Out Common Terms
We can factor out \( 2\sin \alpha \):
\[ z = 2\sin \alpha (\sin \alpha + i \cos \alpha) \]

### Step 4: Recognize the Form of the Complex Number
The expression \( \sin \alpha + i \cos \alpha \) can be rewritten in terms of an exponential form:
\[ \sin \alpha + i \cos \alpha = i(\cos \alpha - i \sin \alpha) = i e^{-i\alpha} \]
Thus, we have:
\[ z = 2\sin \alpha \cdot i e^{-i\alpha} \]

### Step 5: Find the Modulus
The modulus of \( z \) is given by:
\[ |z| = |2\sin \alpha| \cdot |i| \cdot |e^{-i\alpha}| = 2|\sin \alpha| \]
Since \( \alpha \) is in the range \( \frac{3\pi}{2} > \alpha > 2\pi \), \( \sin \alpha \) is negative. Therefore, the modulus is:
\[ |z| = -2\sin \alpha \]

### Step 6: Find the Argument
The argument of \( z \) can be found as follows:
\[ \text{arg}(z) = \text{arg}(2\sin \alpha) + \text{arg}(i) + \text{arg}(e^{-i\alpha}) \]
Since \( \text{arg}(2\sin \alpha) = 0 \) (as it is a positive real number), \( \text{arg}(i) = \frac{\pi}{2} \), and \( \text{arg}(e^{-i\alpha}) = -\alpha \):
\[ \text{arg}(z) = 0 + \frac{\pi}{2} - \alpha = \frac{\pi}{2} - \alpha \]

However, since \( \alpha \) is in the fourth quadrant, we need to adjust the argument:
\[ \text{arg}(z) = \frac{3\pi}{2} - \alpha \]

### Final Results
Thus, the modulus and argument of \( z \) are:
- **Modulus**: \( -2\sin \alpha \)
- **Argument**: \( \frac{3\pi}{2} - \alpha \)