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. $0$
• B. $2i^n$
• C. $2 \cos(n\pi/2)$
• D. $2^n$

Hint:

The squares $(1\pm i)^2 = \pm 2i$ are extremely useful for simplifying powers of these complex numbers.

Answer 1

The Correct Option is C

Step 1: Concept
Simplify the complex terms $(1-i)^2$ and $(1+i)^2$ using algebraic expansion before applying the power $n$.

Step 2: Meaning
Note that $(1-i)^2 = 1 + i^2 - 2i = 1 - 1 - 2i = -2i$ and $(1+i)^2 = 1 + i^2 + 2i = 1 - 1 + 2i = 2i$.

Step 3: Analysis
The expression becomes $\frac{2^n}{(-2i)^n} + \frac{(2i)^n}{2^n} = \frac{1}{(-i)^n} + i^n = i^n + \frac{1}{(-i)^n}$. Using $1/(-i) = i$, this simplifies to $i^n + i^n$ is not correct; rather $i^n + (-i)^{-n} = e^{in\pi/2} + e^{-in\pi/2}$.

Step 4: Conclusion
Using the identity $2 \cos \theta = e^{i\theta} + e^{-i\theta}$, where $\theta = n\pi/2$, the result is $2 \cos(n\pi/2)$. Final Answer: (C)