Question

If \( n \in \mathbb{Z} \), then the expression \[ \frac{2^n}{(1-i)^{2n}} + \frac{(1+i)^{2n}}{2^n} \] is equal to:

• A. \(2(-1)^n\)
• B. \(0\)
• C. \(2\)
• D. \(2^n\)

Hint:

Remember: \[ (1-i)^2=-2i \qquad \text{and} \qquad (1+i)^2=2i \] These identities simplify higher powers quickly.

Answer 1

The Correct Option is A

Concept: Use properties of complex numbers: \[ (1-i)^2 = 1 -2i + i^2 \] and \[ i^2 = -1 \] Simplifying powers of complex numbers reduces the expression easily.

Step 1: Simplify \( (1-i)^2 \). \[ (1-i)^2 = 1 -2i + i^2 \] \[ = 1 -2i -1 \] \[ = -2i \] Thus: \[ (1-i)^{2n} = (-2i)^n \] Hence: \[ \frac{2^n}{(1-i)^{2n}} = \frac{2^n}{(-2i)^n} \] \[ = \frac{1}{(-i)^n} \] \[ = i^n \]

Step 2: Simplify second term. Similarly: \[ (1+i)^2 = 1+2i+i^2 \] \[ = 2i \] Therefore: \[ (1+i)^{2n} = (2i)^n \] Hence: \[ \frac{(1+i)^{2n}}{2^n} = \frac{(2i)^n}{2^n} = i^n \]

Step 3: Add both terms. \[ i^n + i^n = 2i^n \] For integer powers: \[ i^{2k} = (-1)^k \] Thus expression becomes: \[ 2(-1)^n \] \[ \boxed{ 2(-1)^n } \]