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

Jasonbradley

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Jun 16, 2024

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

This problem involves complex numbers and utilizes some key properties to arrive at the solution.

Understanding the Properties

1. Euler's Formula: This formula connects complex exponentials to trigonometric functions: e^(iθ) = cos(θ) + i*sin(θ)

2. De Moivre's Theorem: This theorem states that for any complex number in polar form (r(cos θ + i sin θ)) and any integer n: [r(cos θ + i sin θ)]^n = r^n (cos nθ + i sin nθ)

Applying the Properties

Let's simplify the expression (1+i)^(2024) + (1-i)^(2024).

First, we need to express (1+i) and (1-i) in polar form:

• (1+i):

  • Magnitude: √(1² + 1²) = √2
  • Angle: arctan(1/1) = π/4 radians
  • Polar form: √2(cos(π/4) + i sin(π/4))

• (1-i):

  • Magnitude: √(1² + (-1)²) = √2
  • Angle: arctan(-1/1) = -π/4 radians
  • Polar form: √2(cos(-π/4) + i sin(-π/4))

Now we can apply De Moivre's theorem:

• (1+i)^(2024) = [√2(cos(π/4) + i sin(π/4))]^(2024) = 2^(1012) [cos(2024π/4) + i sin(2024π/4)] = 2^(1012) [cos(506π) + i sin(506π)]

• (1-i)^(2024) = [√2(cos(-π/4) + i sin(-π/4))]^(2024) = 2^(1012) [cos(-2024π/4) + i sin(-2024π/4)] = 2^(1012) [cos(-506π) + i sin(-506π)]

Since the cosine function is even and the sine function is odd, we can simplify further:

• (1+i)^(2024) = 2^(1012) [cos(506π) + i sin(506π)] = 2^(1012) (1 + 0i) = 2^(1012)

• (1-i)^(2024) = 2^(1012) [cos(506π) - i sin(506π)] = 2^(1012) (1 - 0i) = 2^(1012)

Finally, we add the two results:

(1+i)^(2024) + (1-i)^(2024) = 2^(1012) + 2^(1012) = 2^(1013)

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