Question

What is the modulus of the complex number i^{2n + 1}(-i)^{2n - 1}, where n ∈ N and i = √-1?
1. 1
2. - 1
3. √2
4. 2

Answer 1

Correct Answer - Option 1 : 1

Concept:

Iota power:

• i² = -1, i³ = -i, i⁴ = 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)⁴n 

⇒ - (i⁴)n

⇒ - (1)n  = -1

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