What is the value of `( - sqrt( -1) )^(4n+3 ) + ( i^(41) + i^(-257))`, where `n in N `?

Options

A. A)0
B. B)1
C. C)i
D. D)`-i`

Answer

To solve the expression \( (-\sqrt{-1})^{4n+3} + (i^{41} + i^{-257}) \), we will break it down step by step.

### Step 1: Simplify \(-\sqrt{-1}\)
We know that \(-\sqrt{-1} = -i\) since \(i = \sqrt{-1}\). Thus, we can rewrite the expression as:
\[ (-i)^{4n+3} \]

### Step 2: Simplify \((-i)^{4n+3}\)
Using the properties of exponents, we can express \((-i)\) in terms of \(i\):
\[ (-i)^{4n+3} = (-1)^{4n+3} \cdot i^{4n+3} \]
Since \((-1)^{4n+3} = -1\) (because \(4n+3\) is odd), we have:
\[ (-i)^{4n+3} = -i^{4n+3} \]

### Step 3: Simplify \(i^{4n+3}\)
The powers of \(i\) cycle every four terms:
- \(i^1 = i\)
- \(i^2 = -1\)
- \(i^3 = -i\)
- \(i^4 = 1\)

To find \(i^{4n+3}\), we can use the fact that \(4n+3\) modulo 4 gives us the remainder:
\[ 4n + 3 \equiv 3 \mod 4 \]
Thus,
\[ i^{4n+3} = i^3 = -i \]

### Step 4: Substitute back into the expression
Now substituting back, we have:
\[ (-i)^{4n+3} = -(-i) = i \]

### Step 5: Simplify \(i^{41} + i^{-257}\)
Next, we simplify \(i^{41}\) and \(i^{-257}\):
- For \(i^{41}\), we find \(41 \mod 4\):
\[ 41 \equiv 1 \mod 4 \quad \Rightarrow \quad i^{41} = i \]
- For \(i^{-257}\), we first find \(257 \mod 4\):
\[ 257 \equiv 1 \mod 4 \quad \Rightarrow \quad i^{-257} = \frac{1}{i} = -i \]

### Step 6: Combine the results
Now we can combine the results:
\[ i^{41} + i^{-257} = i + (-i) = 0 \]

### Step 7: Final expression
Putting everything together, we have:
\[ (-\sqrt{-1})^{4n+3} + (i^{41} + i^{-257}) = i + 0 = i \]

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