Question

The value of \[ \left[\frac{5i}{(3+i)(3-i)}\right]^{2026} \] is equal to:

• A. \(\frac{1}{2^{2026}}\)
• B. \(\frac{1}{2^{1013}}\)
• C. \(\frac{-1}{2^{1013}}\)
• D. \(\frac{i}{2^{2026}}\)
• E. \(\frac{-1}{2^{2026}}\)

Hint:

To evaluate any high power of \(i\), just look at the last two digits of the exponent. \(26 \pmod 4 = 2\). Thus, \(i^{2026} = i^2 = -1\).

Answer 1

Answer: (E)

Step 1: Understanding the Concept:
We first simplify the complex expression inside the brackets before raising it to the power of 2026.
The denominator is in the form \((a+bi)(a-bi)\).

Step 2: Key Formula or Approach:
1. \((a+bi)(a-bi) = a^2 + b^2\).
2. Powers of \(i\): \(i^1 = i, i^2 = -1, i^3 = -i, i^4 = 1\).

Step 3: Detailed Explanation:
Simplifying the denominator:
\[ (3+i)(3-i) = 3^2 - i^2 = 9 - (-1) = 10 \]
The expression inside the brackets becomes:
\[ \left( \frac{5i}{10} \right) = \frac{i}{2} \]
Raising to the power:
\[ \left( \frac{i}{2} \right)^{2026} = \frac{i^{2026}}{2^{2026}} \]
Evaluating \(i^{2026}\):
Divide 2026 by 4. \(2026 = 4 \times 506 + 2\).
The remainder is 2.
\[ i^{2026} = i^2 = -1 \]
Substituting back:
\[ = \frac{-1}{2^{2026}} \]

Step 4: Final Answer:
The result is \(\frac{-1}{2^{2026}}\).