`[i^(17)+1/(i^(315))]^(9)` is equal to

Options

A. 32i
B. `-512`
C. 512
D. 512i

Answer

To solve the expression \([i^{17} + \frac{1}{i^{315}}]^9\), we will follow these steps:

### Step 1: Simplify \(i^{17}\)
We know that the powers of \(i\) cycle every 4:
- \(i^1 = i\)
- \(i^2 = -1\)
- \(i^3 = -i\)
- \(i^4 = 1\)

To find \(i^{17}\), we can reduce the exponent modulo 4:
\[ 17 \mod 4 = 1 \quad \Rightarrow \quad i^{17} = i \]

### Step 2: Simplify \(\frac{1}{i^{315}}\)
Similarly, we simplify \(i^{315}\) by reducing the exponent modulo 4:
\[ 315 \mod 4 = 3 \quad \Rightarrow \quad i^{315} = -i \]
Thus,
\[ \frac{1}{i^{315}} = \frac{1}{-i} = -\frac{1}{i} = -(-i) = i \]

### Step 3: Combine the results
Now we can substitute back into the expression:
\[ i^{17} + \frac{1}{i^{315}} = i + i = 2i \]

### Step 4: Raise to the power of 9
Now we raise the result to the power of 9:
\[ (2i)^9 = 2^9 \cdot i^9 \]

Calculating \(2^9\):
\[ 2^9 = 512 \]

Next, we simplify \(i^9\):
\[ 9 \mod 4 = 1 \quad \Rightarrow \quad i^9 = i \]

### Step 5: Final result
Putting it all together:
\[ (2i)^9 = 512 \cdot i \]

Thus, the final answer is:
\[ \boxed{512i} \]