If `z=re^(itheta)`, then `|e^(iz)|` is equal to

Options

A. `e^(-rcostheta)`
B. `e^(rcostheta)`
C. `e^(rsintheta)`
D. `e^(-rsintheta)`

Explanation

To solve the problem, we need to find the value of \( |e^{iz}| \) given that \( z = re^{i\theta} \).

### Step-by-Step Solution:

1. **Substitute for \( z \)**:
We start with the expression for \( z \):
\[ z = re^{i\theta} \]
Now, we need to find \( e^{iz} \):
\[ e^{iz} = e^{i(re^{i\theta})} \]

2. **Express \( e^{iz} \)**:
We can simplify \( e^{iz} \):
\[ e^{iz} = e^{i(r(\cos \theta + i \sin \theta))} = e^{i(r \cos \theta + ir \sin \theta)} = e^{-r \sin \theta} e^{i r \cos \theta} \]
Here, we used the property \( e^{a + b} = e^a e^b \).

3. **Find the modulus**:
Now, we need to find the modulus of \( e^{iz} \):
\[ |e^{iz}| = |e^{-r \sin \theta} e^{i r \cos \theta}| \]
Using the property of modulus that states \( |ab| = |a||b| \):
\[ |e^{iz}| = |e^{-r \sin \theta}| \cdot |e^{i r \cos \theta}| \]

4. **Calculate each modulus**:
- The modulus of \( e^{-r \sin \theta} \) is simply \( e^{-r \sin \theta} \) since it is a real number.
- The modulus of \( e^{i r \cos \theta} \) is \( 1 \) because the modulus of any complex exponential \( e^{i\phi} \) is \( 1 \).

5. **Combine the results**:
Therefore, we have:
\[ |e^{iz}| = e^{-r \sin \theta} \cdot 1 = e^{-r \sin \theta} \]

### Final Result:
Thus, the value of \( |e^{iz}| \) is:
\[ \boxed{e^{-r \sin \theta}} \]