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. $2i^n$
• B. 0
• C. $2^{n+1}$
• D. $i^n$

Hint:

Remember that $(1+i)^2 = 2i$ and $(1-i)^2 = -2i$. These are very common shortcuts in complex number problems.

Answer 1

The Correct Option is A

Step 1: Concept
Simplify the complex terms $(1-i)^2$ and $(1+i)^2$ first.

Step 2: Meaning
$(1-i)^2 = 1 - 2i + i^2 = 1 - 2i - 1 = -2i$
$(1+i)^2 = 1 + 2i + i^2 = 1 + 2i - 1 = 2i$

Step 3: Analysis
Substitute these into the expression:
$\frac{2^n}{(-2i)^n} + \frac{(2i)^n}{2^n} = \frac{2^n}{(-2)^n i^n} + \frac{2^n i^n}{2^n}$
$= \frac{1}{(-i)^n} + i^n$
Since $\frac{1}{-i} = i$, then $(\frac{1}{-i})^n = i^n$.
So, $i^n + i^n = 2i^n$.

Step 4: Conclusion
The expression simplifies to $2i^n$. Final Answer: (A)