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

Options

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

Answer

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

### Step-by-Step Solution:

1. **Substituting for \( z \)**:
We start with the expression for \( z \):
\[ z = re^{i\theta} \]
Therefore, we can express \( iz \) as:
\[ iz = i(re^{i\theta}) = r(i e^{i\theta}) \]

2. **Using Euler's Formula**:
We can rewrite \( e^{i\theta} \) using Euler's formula:
\[ e^{i\theta} = \cos(\theta) + i\sin(\theta) \]
Thus, we have:
\[ iz = r(i(\cos(\theta) + i\sin(\theta))) = r(i\cos(\theta) - \sin(\theta)) \]

3. **Finding \( e^{iz} \)**:
Now we need to find \( e^{iz} \):
\[ e^{iz} = e^{r(i\cos(\theta) - \sin(\theta))} \]
This can be separated into:
\[ e^{iz} = e^{r(-\sin(\theta))} \cdot e^{ir\cos(\theta)} \]

4. **Taking the Modulus**:
We want to find the modulus \( |e^{iz}| \):
\[ |e^{iz}| = |e^{r(-\sin(\theta))} \cdot e^{ir\cos(\theta)}| \]
The modulus of a product is the product of the moduli:
\[ |e^{iz}| = |e^{r(-\sin(\theta))}| \cdot |e^{ir\cos(\theta)}| \]

5. **Calculating Each Modulus**:
- The modulus of \( e^{r(-\sin(\theta))} \) is:
\[ |e^{r(-\sin(\theta))}| = e^{r(-\sin(\theta))} \]
- The modulus of \( e^{ir\cos(\theta)} \) is:
\[ |e^{ir\cos(\theta)}| = 1 \]
(since \( e^{i\text{anything}} \) has a modulus of 1).

6. **Final Result**:
Therefore, we have:
\[ |e^{iz}| = e^{-r\sin(\theta)} \cdot 1 = e^{-r\sin(\theta)} \]

### Conclusion:
The final answer is:
\[ |e^{iz}| = e^{-r\sin(\theta)} \]