`i^(-35)` is equal to:

Options

A. `i`
B. `-i`
C. `1`
D. `-1`

Answer

To solve the problem \( i^{-35} \), we can follow these steps:

### Step 1: Rewrite the expression
We start by rewriting \( i^{-35} \) using the property of exponents:
\[ i^{-35} = \frac{1}{i^{35}} \]

### Step 2: Find \( i^{35} \)
Next, we need to find \( i^{35} \). The powers of \( i \) cycle every 4:
- \( i^1 = i \)
- \( i^2 = -1 \)
- \( i^3 = -i \)
- \( i^4 = 1 \)

To find \( i^{35} \), we can use the modulo operation:
\[ 35 \mod 4 \]
Calculating \( 35 \div 4 \) gives a quotient of 8 and a remainder of 3. Thus:
\[ 35 \mod 4 = 3 \]
This means:
\[ i^{35} = i^3 \]

### Step 3: Evaluate \( i^3 \)
From our earlier calculations:
\[ i^3 = -i \]

### Step 4: Substitute back into the expression
Now we substitute \( i^{35} \) back into the expression for \( i^{-35} \):
\[ i^{-35} = \frac{1}{i^{35}} = \frac{1}{-i} \]

### Step 5: Simplify \( \frac{1}{-i} \)
To simplify \( \frac{1}{-i} \), we can multiply the numerator and denominator by \( i \) to eliminate the imaginary unit from the denominator:
\[ \frac{1}{-i} \cdot \frac{i}{i} = \frac{i}{-i^2} \]
Since \( i^2 = -1 \), we have:
\[ -i^2 = -(-1) = 1 \]
Thus:
\[ \frac{i}{-i^2} = \frac{i}{1} = i \]

### Final Answer
Therefore, the value of \( i^{-35} \) is:
\[ \boxed{i} \]

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