If ` a, b, in R " then " |e^(a+ ib)|`

Options

A. `e^(a)`
B. `e^(b)`
C. 1
D. None

Answer

To find the modulus of the complex number \( e^{a + ib} \), where \( a, b \in \mathbb{R} \), we can follow these steps:

### Step-by-Step Solution:

1. **Express the complex number**:
We start with the expression \( e^{a + ib} \). According to the properties of exponents, we can separate the real and imaginary parts:
\[ e^{a + ib} = e^a \cdot e^{ib} \]

2. **Use Euler's formula**:
Euler's formula states that \( e^{ib} = \cos(b) + i\sin(b) \). Thus, we can rewrite our expression as:
\[ e^{a + ib} = e^a \cdot (\cos(b) + i\sin(b)) \]

3. **Find the modulus**:
The modulus of a complex number \( z = x + iy \) is given by \( |z| = \sqrt{x^2 + y^2} \). In our case, we have:
\[ |e^{a + ib}| = |e^a \cdot (\cos(b) + i\sin(b))| \]

By the property of moduli, we can separate the modulus:
\[ |e^{a + ib}| = |e^a| \cdot |(\cos(b) + i\sin(b))| \]

4. **Calculate the modulus of each part**:
- The modulus of \( e^a \) is simply \( e^a \) since \( a \) is a real number.
- The modulus of \( \cos(b) + i\sin(b) \) can be calculated as:
\[ |(\cos(b) + i\sin(b))| = \sqrt{\cos^2(b) + \sin^2(b)} \]
Using the Pythagorean identity \( \cos^2(b) + \sin^2(b) = 1 \), we find:
\[ |(\cos(b) + i\sin(b))| = \sqrt{1} = 1 \]

5. **Combine the results**:
Now, we can combine the results from the modulus calculations:
\[ |e^{a + ib}| = e^a \cdot 1 = e^a \]

### Final Answer:
Thus, the modulus \( |e^{a + ib}| \) is:
\[ \boxed{e^a} \]