Correct Answer - Option 1 : 1

**Concept:**

**Iota power:**

- i
^{2} = -1, i^{3} = -i, i^{4} = 1
- Number of the form 2n is always even, n ∈ N
- Numebr of the form 2n +1 or 2n - 1 is always odd, n ∈ N
- (-a)
^{2n -1 }= -(a)^{2n -1}

**Calculation:**

Given:

i2n + 1(-i)2n - 1

From the concept used we know that 2n - 1 is odd

⇒ i2n + 1(-i)2n - 1

⇒ - (i)2n + 1(i)2n - 1

⇒ - (i)2n + 1 + 2n - 1

⇒ - (i)^{4}n

⇒ - (i^{4})n

⇒ - (1)n = -1

**Hence, the modulus of the complex number i2n + 1(-i)**^{2n - 1} is 1.