What is the modulus of the complex number `i^(2n+1) (-i)^(2n-1)`, where `n in N` and `i=sqrt(-1)`?

Options

A. `-1`
B. 1
C. `sqrt(2)`
D. 2

Answer

To find the modulus of the complex number \( i^{2n+1} (-i)^{2n-1} \), where \( n \in \mathbb{N} \) and \( i = \sqrt{-1} \), we can follow these steps:

### Step 1: Simplify the expression
We start with the expression:
\[ i^{2n+1} (-i)^{2n-1} \]
We can rewrite \( -i \) as \( i^3 \) (since \( -i = i^{3} \)). Thus, we have:
\[ (-i)^{2n-1} = (i^3)^{2n-1} = i^{3(2n-1)} = i^{6n-3} \]
Now, substituting this back into our expression gives:
\[ i^{2n+1} \cdot i^{6n-3} \]

### Step 2: Combine the powers of \( i \)
Using the property of exponents \( a^m \cdot a^n = a^{m+n} \), we combine the powers:
\[ i^{(2n+1) + (6n-3)} = i^{8n - 2} \]

### Step 3: Find the modulus of \( i^{8n - 2} \)
The modulus of any complex number \( z = re^{i\theta} \) is given by \( |z| = r \). For any power of \( i \), we know:
- \( i^0 = 1 \) (modulus = 1)
- \( i^1 = i \) (modulus = 1)
- \( i^2 = -1 \) (modulus = 1)
- \( i^3 = -i \) (modulus = 1)
- \( i^4 = 1 \) (modulus = 1), and so on.

Thus, the modulus of \( i^k \) for any integer \( k \) is always 1. Therefore:
\[ |i^{8n - 2}| = 1 \]

### Conclusion
The modulus of the complex number \( i^{2n+1} (-i)^{2n-1} \) is:
\[ \boxed{1} \]