If `(1+i^2 +i^4 +i^6 +i^208)=a +ib`,then the value of (a,b) is

Options

A. `-1`
B. 1
C. 0
D. `pm 1`

Answer

To solve the expression \(1 + i^2 + i^4 + i^6 + i^{208}\) and express it in the form \(a + ib\), we can follow these steps:

### Step 1: Identify the powers of \(i\)
Recall that \(i\) is the imaginary unit defined as \(i = \sqrt{-1}\). The powers of \(i\) cycle every four terms:
- \(i^0 = 1\)
- \(i^1 = i\)
- \(i^2 = -1\)
- \(i^3 = -i\)
- \(i^4 = 1\) (and the cycle repeats)

### Step 2: Calculate each term
1. **Calculate \(i^2\)**:
\[ i^2 = -1 \]

2. **Calculate \(i^4\)**:
\[ i^4 = 1 \]

3. **Calculate \(i^6\)**:
Since \(i^6 = i^{4+2} = i^4 \cdot i^2 = 1 \cdot (-1) = -1\).

4. **Calculate \(i^{208}\)**:
To find \(i^{208}\), we can reduce the exponent modulo 4:
\[ 208 \mod 4 = 0 \quad \text{(since } 208 = 4 \times 52\text{)} \]
Thus, \(i^{208} = i^0 = 1\).

### Step 3: Substitute back into the expression
Now substitute the calculated values back into the expression:
\[ 1 + i^2 + i^4 + i^6 + i^{208} = 1 + (-1) + 1 + (-1) + 1 \]

### Step 4: Simplify the expression
Now simplify the expression:
\[ 1 - 1 + 1 - 1 + 1 = 1 \]

### Step 5: Express in the form \(a + ib\)
The result \(1\) can be expressed as:
\[ 1 + 0i \]
Thus, we have \(a = 1\) and \(b = 0\).

### Final Answer
The values of \(a\) and \(b\) are:
\[ (a, b) = (1, 0) \]