Question

If $\alpha$ is a real number satisfying $\alpha^2 - \frac{1}{\alpha^2} = 2$, then the value of $(\alpha + \frac{i}{\alpha})^{16}$ is equal to

• A. 2048
• B. 4096
• C. 2024
• D. 4048
• E. 5096

Hint:

For expressions like $(1 \pm i)^n$, always reduce the power by squaring first. Since $(1+i)^2 = 2i$ and $(1-i)^2 = -2i$, high powers become simple powers of 2.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The problem relates a real variable algebraic condition to a high power of a complex expression. Squaring the base of the power will likely simplify the expression using the given condition.

Step 2: Detailed Explanation:
Let \( w = \alpha + \frac{i}{\alpha} \).
Evaluate \( w^2 \):
\[ w^2 = (\alpha + \frac{i}{\alpha})^2 = \alpha^2 + (\frac{i}{\alpha})^2 + 2(\alpha)(\frac{i}{\alpha}) \]
\[ w^2 = \alpha^2 + \frac{i^2}{\alpha^2} + 2i \]
\[ w^2 = \alpha^2 - \frac{1}{\alpha^2} + 2i \]
Using the given condition \( \alpha^2 - \frac{1}{\alpha^2} = 2 \):
\[ w^2 = 2 + 2i \]
We need to find \( w^{16} \), which can be written as \( (w^2)^8 \):
\[ w^{16} = (2 + 2i)^8 = [2(1+i)]^8 = 2^8 \cdot (1+i)^8 \]
Recall that \( (1+i)^2 = 2i \). Thus:
\[ (1+i)^8 = ((1+i)^2)^4 = (2i)^4 = 2^4 \cdot i^4 = 16 \cdot 1 = 16 \]
Finally:
\[ w^{16} = 2^8 \cdot 16 = 256 \cdot 16 = 4096 \]

Step 3: Final Answer:
The value is 4096.