`[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 \(i^{17}\)
Recall that \(i\) is the imaginary unit, defined as \(i = \sqrt{-1}\). The powers of \(i\) cycle every four:
- \(i^1 = i\)
- \(i^2 = -1\)
- \(i^3 = -i\)
- \(i^4 = 1\)

To simplify \(i^{17}\), we can find the equivalent power within the cycle:
\[ 17 \mod 4 = 1 \]
Thus, \(i^{17} = i\).

### Step 2: Simplify \(\frac{1}{i^{315}}\)
Next, we simplify \(i^{315}\) using the same cycle:
\[ 315 \mod 4 = 3 \]
So, \(i^{315} = -i\). Now, we can find \(\frac{1}{i^{315}}\):
\[ \frac{1}{i^{315}} = \frac{1}{-i} = -\frac{1}{i} \]
To simplify \(-\frac{1}{i}\), we can multiply the numerator and denominator by \(i\):
\[ -\frac{1}{i} = -\frac{i}{i^2} = -\frac{i}{-1} = i \]

### Step 3: Combine the results
Now we can combine the results from Steps 1 and 2:
\[ i^{17} + \frac{1}{i^{315}} = i + i = 2i \]

### Step 4: Raise to the power of 9
Now we need to raise \(2i\) to the power of 9:
\[ (2i)^9 = 2^9 \cdot i^9 \]
Calculating \(2^9\):
\[ 2^9 = 512 \]
Next, we simplify \(i^9\):
\[ 9 \mod 4 = 1 \]
Thus, \(i^9 = i\).

### Step 5: Final result
Combining these results gives:
\[ (2i)^9 = 512 \cdot i \]

### Conclusion
The final answer is:
\[ \boxed{512i} \]