`(-i)^(4n+3)`, where n Is a positive integer. Then prove that it is equals to i

Explanation

To prove that \((-i)^{4n+3} = i\) where \(n\) is a positive integer, we will follow these steps:

### Step 1: Rewrite the expression
We start with the expression:
\[ (-i)^{4n+3} \]

### Step 2: Break down the exponent
We can rewrite the exponent \(4n + 3\) as:
\[ (-i)^{4n + 3} = (-i)^{4n} \cdot (-i)^{3} \]

### Step 3: Simplify \((-i)^{4n}\)
Using the property of exponents, we know that:
\[ (-i)^{4n} = ((-i)^4)^n \]
Now, we calculate \((-i)^4\):
\[ (-i)^4 = (-1)^4 \cdot i^4 = 1 \cdot 1 = 1 \]
Thus, we have:
\[ (-i)^{4n} = 1^n = 1 \]

### Step 4: Simplify \((-i)^{3}\)
Next, we calculate \((-i)^{3}\):
\[ (-i)^{3} = -i \cdot -i \cdot -i = -(-1) \cdot i^3 = -(-1) \cdot (-i) = -(-i) = i \]

### Step 5: Combine the results
Now we can combine the results from Steps 3 and 4:
\[ (-i)^{4n + 3} = (-i)^{4n} \cdot (-i)^{3} = 1 \cdot i = i \]

### Conclusion
Thus, we have shown that:
\[ (-i)^{4n + 3} = i \]
This completes the proof.