(1-i)^2024

3 min read Jun 16, 2024
(1-i)^2024

Exploring the Power of Complex Numbers: (1-i)^2024

The expression (1-i)^2024 might seem intimidating at first glance, especially with the large exponent. However, with a bit of understanding about complex numbers and their properties, we can simplify this expression and find its value.

Understanding Complex Numbers

Complex numbers are numbers that can be expressed in the form a + bi, where a and b are real numbers and i is the imaginary unit, defined as the square root of -1.

Key Properties:

  • Modulus: The modulus of a complex number a + bi is denoted as |a + bi| and is calculated as √(a² + b²).
  • Argument: The argument of a complex number a + bi is the angle it makes with the positive real axis in the complex plane. It's denoted as arg(a + bi).
  • De Moivre's Theorem: This theorem 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 (1-i)^2024

  1. Expressing in Polar Form: First, we need to express (1-i) in polar form. Its modulus is √(1² + (-1)²) = √2, and its argument is -π/4 (since it lies in the fourth quadrant). Therefore, (1-i) = √2 (cos (-π/4) + i sin (-π/4)).

  2. Applying De Moivre's Theorem: Using De Moivre's theorem, we can calculate (1-i)^2024 as:

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

  1. Simplifying: Since 2024 is divisible by 4, the angle -π/4 * 2024 simplifies to a multiple of 2π, which means the cosine term is 1 and the sine term is 0. Additionally, (√2)^2024 = 2^1012.

Therefore, (1-i)^2024 = 2^1012 (1 + 0i) = 2^1012

Conclusion

The seemingly complex expression (1-i)^2024 simplifies to a real number, 2^1012, by using the properties of complex numbers and De Moivre's theorem. This example highlights the power of using these tools to solve problems that may seem daunting at first.

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