Question

The value of \[ \left(\frac{10i}{(2-i)(3-i)}\right)^{2024} \] is equal to:

• A. \(2^{2024}\)
• B. \(2^{1012}\)
• C. \(4^{2024}\)
• D. \(\left(\frac{1}{2}\right)^{2024}\)
• E. \(\left(\frac{1}{2}\right)^{1012}\)

Hint:

Use polar form and Euler's formula for powers of complex numbers to simplify calculations, especially with high powers or complex angles.

Answer 1

The Correct Option is B

First, simplify the expression inside the parentheses: \[ \frac{10i}{(2-i)(3-i)} \] Compute the product in the denominator: \[ (2-i)(3-i) = 6 - 5i + i^2 = 6 - 5i - 1 = 5 - 5i \] Thus, the expression becomes: \[ \frac{10i}{5 - 5i} \] Simplify this by multiplying the numerator and the denominator by the conjugate of the denominator: \[ \frac{10i}{5 - 5i} \cdot \frac{5 + 5i}{5 + 5i} = \frac{50i + 50i^2}{25 + 25i - 25i - 25i^2} = \frac{50i - 50}{25 + 25} = \frac{-50 + 50i}{50} \] \[ = -1 + i \] Now, we raise \((-1+i)\) to the power of 2024: \[ (-1+i)^{2024} = (i-1)^{2024} \] By Euler's formula, express \(i-1\) in polar form: \[ i-1 = \sqrt{2} e^{i\left(\frac{3\pi}{4}\right)} \] Raise to the 2024th power: \[ \left(\sqrt{2} e^{i\left(\frac{3\pi}{4}\right)}\right)^{2024} = 2^{1012} e^{i\left(\frac{3\pi}{4} \times 2024\right)} \] Calculate the angle modulo \(2\pi\): \[ \frac{3\pi \times 2024}{4} \mod 2\pi = 0 \] Hence, the expression simplifies to: \[ 2^{1012} \] Thus, the correct answer is option (B), \(2^{1012}\).