(1+i)^2024+(1-i)^2024

Jasonbradley

3 min read

·

Jun 16, 2024

21

Share

Evaluating (1 + i)^2024 + (1 - i)^2024

This article explores the evaluation of the expression (1 + i)^2024 + (1 - i)^2024, where i represents the imaginary unit (√-1).

Understanding Complex Numbers and De Moivre's Theorem

Complex numbers are numbers of the form a + bi, where a and b are real numbers and i is the imaginary unit. They can be represented geometrically as points in the complex plane.

De Moivre's Theorem is a powerful tool for calculating powers of complex numbers. It states that for any complex number z = r(cos θ + i sin θ) and any integer n,

(z)^n = r^n(cos(nθ) + i sin(nθ))

Simplifying the Expression

Let's begin by simplifying the expression (1 + i)^2024 + (1 - i)^2024.

1. Finding the Modulus and Argument:

• Modulus: The modulus of a complex number is its distance from the origin in the complex plane. The modulus of both (1 + i) and (1 - i) is √(1² + 1²) = √2.

• Argument: The argument of a complex number is the angle it makes with the positive real axis. The argument of (1 + i) is π/4, and the argument of (1 - i) is -π/4.

2. Applying De Moivre's Theorem:

• (1 + i)^2024: Using De Moivre's Theorem, we have:

(1 + i)^2024 = (√2)^2024(cos(2024π/4) + i sin(2024π/4))

• (1 - i)^2024: Similarly, we have:

(1 - i)^2024 = (√2)^2024(cos(-2024π/4) + i sin(-2024π/4))

3. Evaluating the Trigonometric Functions:

• cos(2024π/4) = cos(506π) = 1
• sin(2024π/4) = sin(506π) = 0
• cos(-2024π/4) = cos(-506π) = 1
• sin(-2024π/4) = sin(-506π) = 0

4. Simplifying the Expression:

Now, substituting these values back into the expression, we get:

(1 + i)^2024 + (1 - i)^2024 = (√2)^2024(1 + 0) + (√2)^2024(1 + 0)

5. Final Result:

Finally, simplifying the expression, we get:

(1 + i)^2024 + (1 - i)^2024 = 2 * (√2)^2024 = 2^(1013)

Therefore, the value of the expression (1 + i)^2024 + (1 - i)^2024 is 2^(1013).