Question

Evaluate the following expression: \[ (1+i)^{2024} + (1-i)^{2024} \]

• A. \( -2^{1012} \)
• B. \( 2^{1013} \)
• C. \( 2^{2024}i \)
• D. \( -2^{1012}i \)

Hint:

Use De Moivre’s Theorem to simplify powers of complex numbers. Calculate arguments carefully to simplify the result.

Answer 1

The Correct Option is A

We are given the expression \( (1+i)^{2024} + (1-i)^{2024} \). Let’s first express \( 1+i \) and \( 1-i \) in polar form.
Step 1: Express \( 1+i \) and \( 1-i \) in polar form. For \( 1+i \), the modulus is \( |1+i| = \sqrt{2} \), and the argument is \( \arg(1+i) = \frac{\pi}{4} \). Thus: \[ 1+i = \sqrt{2} \left( \cos\frac{\pi}{4} + i\sin\frac{\pi}{4} \right) \]
Similarly, for \( 1-i \), the modulus is also \( \sqrt{2} \), and the argument is \( \arg(1-i) = -\frac{\pi}{4} \). Thus: \[ 1-i = \sqrt{2} \left( \cos\left(-\frac{\pi}{4}\right) + i\sin\left(-\frac{\pi}{4}\right) \right) \]
Step 2: Apply De Moivre’s Theorem. Now we apply De Moivre’s Theorem to calculate the powers: \[ (1+i)^{2024} = \left(\sqrt{2}\right)^{2024} \left( \cos\left(2024 \times \frac{\pi}{4}\right) + i\sin\left(2024 \times \frac{\pi}{4}\right) \right) \] \[ (1-i)^{2024} = \left(\sqrt{2}\right)^{2024} \left( \cos\left(2024 \times -\frac{\pi}{4}\right) + i\sin\left(2024 \times -\frac{\pi}{4}\right) \right) \]
Step 3: Simplify the expression. Since \( \left(\sqrt{2}\right)^{2024} = 2^{1012} \), and using the periodicity of sine and cosine, both arguments will be equivalent to 0 modulo \( 2\pi \). Therefore, the sum is: \[ 2^{1012} \times 2 = 2^{1013} \] Thus, the correct answer is option (1).