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:
Evaluating a high power of a complex expression can be simplified by expanding the square of the base first.
We use the given algebraic condition for \(\alpha\).
Step 2: Key Formula or Approach:
Use \((x + y)^{16} = \left((x + y)^2\right)^8\).
Expand the square and substitute \(\alpha^2 - \frac{1}{\alpha^2} = 2\).
Step 3: Detailed Explanation:
Square the base:
\[ \left(\alpha + \frac{i}{\alpha}\right)^2 = \alpha^2 + 2(\alpha)\left(\frac{i}{\alpha}\right) + \left(\frac{i}{\alpha}\right)^2 \] \[ = \alpha^2 + 2i + \frac{i^2}{\alpha^2} = \alpha^2 - \frac{1}{\alpha^2} + 2i \] Substitute the given condition:
\[ = 2 + 2i \] Now raise this to the 8th power:
\[ \left(\alpha + \frac{i}{\alpha}\right)^{16} = (2 + 2i)^8 = [2(1 + i)]^8 = 2^8(1 + i)^8 \] We know \((1 + i)^2 = 2i\).
So, \((1 + i)^8 = ((1 + i)^2)^4 = (2i)^4 = 16i^4 = 16\).
Multiply everything:
\[ 2^8 \cdot 16 = 256 \cdot 16 = 4096 \] Step 4: Final Answer:
The value is 4096.