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: Understanding the Concept:
The problem involves powers of complex numbers. Simplifying terms like \((1 \pm i)^2\) often leads to simpler imaginary units.
Step 2: Detailed Explanation:
Evaluate the basic squares first:
\[ (1-i)^2 = 1 - 2i + i^2 = 1 - 2i - 1 = -2i \]
\[ (1+i)^2 = 1 + 2i + i^2 = 1 + 2i - 1 = 2i \]

Now, substitute these back into the given expression:
\[ \frac{2^n}{((1-i)^2)^n} + \frac{((1+i)^2)^n}{2^n} \]
\[ = \frac{2^n}{(-2i)^n} + \frac{(2i)^n}{2^n} \]
\[ = \frac{2^n}{(-2)^n \cdot i^n} + \frac{2^n \cdot i^n}{2^n} \]
\[ = \frac{1}{(-1)^n \cdot i^n} + i^n \]
Note that \(\frac{1}{i} = -i\). So \(\frac{1}{i^n} = (-i)^n\).
\[ = \frac{(-i)^n}{(-1)^n} + i^n \]
\[ = \left(\frac{-i}{-1}\right)^n + i^n \]
\[ = i^n + i^n = 2i^n \]
Step 3: Final Answer:
The expression simplifies to \(2i^n\).