If `pi/2` < `alpha` < `3pi`/2 then the modulus argument of `(1+cos 2alpha)+i sin2alpha`

Options

A. `-2cosalpha(cos(pi+alpha)+isin(pi+alpha))`
B. `2cosalpha(cosalpha+isinalpha)`
C. `2cosalpha(cos(-alpha)+isin(-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 \( \frac{\pi}{2} < \alpha < \frac{3\pi}{2} \), we can follow these steps:

### Step 1: Rewrite the expression
We start with the expression:
\[ z = (1 + \cos 2\alpha) + i \sin 2\alpha \]

### Step 2: Use trigonometric identities
Using the double angle identities:
\[ \cos 2\alpha = 2\cos^2 \alpha - 1 \quad \text{or} \quad \cos 2\alpha = 1 - 2\sin^2 \alpha \]
We can rewrite \( 1 + \cos 2\alpha \):
\[ 1 + \cos 2\alpha = 1 + (2\cos^2 \alpha - 1) = 2\cos^2 \alpha \]
Thus, we can express \( z \) as:
\[ z = 2\cos^2 \alpha + i \sin 2\alpha \]

### Step 3: Rewrite \( \sin 2\alpha \)
Using the double angle identity for sine:
\[ \sin 2\alpha = 2\sin \alpha \cos \alpha \]
Now we can write:
\[ z = 2\cos^2 \alpha + i(2\sin \alpha \cos \alpha) \]

### Step 4: Factor out common terms
We can factor out \( 2\cos \alpha \):
\[ z = 2\cos \alpha (\cos \alpha + i \sin \alpha) \]
Recognizing that \( \cos \alpha + i \sin \alpha = e^{i\alpha} \), we can write:
\[ z = 2\cos \alpha \cdot e^{i\alpha} \]

### Step 5: Find the modulus
The modulus of \( z \) is given by:
\[ |z| = |2\cos \alpha| \cdot |e^{i\alpha}| = 2|\cos \alpha| \cdot 1 = 2|\cos \alpha| \]

### Step 6: Find the argument
The argument of \( z \) is:
\[ \arg(z) = \arg(2\cos \alpha) + \arg(e^{i\alpha}) = \arg(2\cos \alpha) + \alpha \]
Since \( 2\cos \alpha \) is real and its sign depends on \( \alpha \):
- If \( \frac{\pi}{2} < \alpha < \frac{3\pi}{2} \), then \( \cos \alpha < 0 \) and thus \( \arg(2\cos \alpha) = \pi \).

Therefore, the argument becomes:
\[ \arg(z) = \pi + \alpha \]

### Final Result
Thus, the modulus and argument of \( z \) are:
\[ |z| = 2|\cos \alpha| \quad \text{and} \quad \arg(z) = \pi + \alpha \]