the value of `((i^(11)+i^(12)+i^(13)+i^(14) +i^(15)))/((1+i))`is

Options

A. `(-(1+i))/2`
B. `((1-i))/2`
C. ` ((1+i))/2`
D. `1/2`

Answer

To solve the expression \(\frac{i^{11} + i^{12} + i^{13} + i^{14} + i^{15}}{1 + i}\), we will follow these steps:

### Step 1: Simplify the powers of \(i\)

Recall that the powers of \(i\) cycle every 4 terms:
- \(i^1 = i\)
- \(i^2 = -1\)
- \(i^3 = -i\)
- \(i^4 = 1\)

Using this, we can simplify each term in the numerator:

- \(i^{11} = i^{4 \cdot 2 + 3} = i^3 = -i\)
- \(i^{12} = i^{4 \cdot 3} = 1\)
- \(i^{13} = i^{4 \cdot 3 + 1} = i^1 = i\)
- \(i^{14} = i^{4 \cdot 3 + 2} = i^2 = -1\)
- \(i^{15} = i^{4 \cdot 3 + 3} = i^3 = -i\)

### Step 2: Substitute the simplified values into the expression

Now substitute these values back into the expression:

\[ i^{11} + i^{12} + i^{13} + i^{14} + i^{15} = (-i) + 1 + i + (-1) + (-i) \]

### Step 3: Combine like terms

Now, combine the terms:

\[ (-i + i - i) + (1 - 1) = -i + 0 = -i \]

### Step 4: Write the expression with the simplified numerator

Now we have:

\[ \frac{-i}{1 + i} \]

### Step 5: Rationalize the denominator

To simplify \(\frac{-i}{1 + i}\), we multiply the numerator and the denominator by the conjugate of the denominator, which is \(1 - i\):

\[ \frac{-i(1 - i)}{(1 + i)(1 - i)} \]

Calculating the denominator:

\[ (1 + i)(1 - i) = 1^2 - i^2 = 1 - (-1) = 2 \]

Calculating the numerator:

\[ -i(1 - i) = -i + i^2 = -i - 1 \]

So we have:

\[ \frac{-1 - i}{2} \]

### Step 6: Write the final expression

Thus, we can express the final result as:

\[ \frac{-1}{2} - \frac{i}{2} \]

This can also be written as:

\[ \frac{1 - i}{2} \]

### Final Answer

The value of the expression is:

\[ \frac{1 - i}{2} \]