If multiplicative inverse of (1 + i) is a + ib then (a + b) equals to

Answer

To find the value of \( a + b \) where the multiplicative inverse of \( 1 + i \) is expressed as \( a + ib \), we can follow these steps:

### Step 1: Identify the multiplicative inverse
The multiplicative inverse of a complex number \( z = x + iy \) is given by:
\[ \frac{1}{z} = \frac{1}{x + iy} \]
In our case, \( z = 1 + i \).

### Step 2: Multiply by the conjugate
To find the multiplicative inverse, we multiply the numerator and the denominator by the conjugate of the denominator:
\[ \frac{1}{1 + i} \cdot \frac{1 - i}{1 - i} = \frac{1 - i}{(1 + i)(1 - i)} \]

### Step 3: Simplify the denominator
The denominator simplifies as follows:
\[ (1 + i)(1 - i) = 1^2 - i^2 = 1 - (-1) = 1 + 1 = 2 \]

### Step 4: Write the expression for the inverse
Now we can write the expression for the multiplicative inverse:
\[ \frac{1 - i}{2} = \frac{1}{2} - \frac{i}{2} \]

### Step 5: Identify \( a \) and \( b \)
From the expression \( \frac{1}{2} - \frac{i}{2} \), we can identify:
\[ a = \frac{1}{2}, \quad b = -\frac{1}{2} \]

### Step 6: Calculate \( a + b \)
Now we can find \( a + b \):
\[ a + b = \frac{1}{2} + \left(-\frac{1}{2}\right) = \frac{1}{2} - \frac{1}{2} = 0 \]

### Final Answer
Thus, the value of \( a + b \) is:
\[ \boxed{0} \]