%5E2000.jpeg "(1+i)^2000")
Exploring the Power of Complex Numbers: (1 + i)^2000
The expression (1 + i)^2000 presents an intriguing challenge in the world of complex numbers. While it may seem daunting at first glance, understanding the properties of complex numbers and De Moivre's Theorem allows us to solve this problem efficiently.
De Moivre's Theorem: The Key to Complex Power
De Moivre's Theorem states that for any complex number in polar form, z = r(cos θ + i sin θ), and any integer n:
z^n = r^n(cos(nθ) + i sin(nθ))
This theorem provides a powerful way to calculate powers of complex numbers.
Solving (1 + i)^2000
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Convert to Polar Form: First, we need to express (1 + i) in polar form.
- Magnitude (r): |1 + i| = √(1² + 1²) = √2
- Angle (θ): arctan(1/1) = π/4 radians
Therefore, (1 + i) = √2(cos(π/4) + i sin(π/4))
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Apply De Moivre's Theorem:
(1 + i)^2000 = [√2(cos(π/4) + i sin(π/4))]^2000 = (√2)^2000 * (cos(2000 * π/4) + i sin(2000 * π/4))
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Simplify:
- (√2)^2000 = 2^1000
- 2000 * π/4 = 500π
Since 500π is a multiple of 2π, we know that:
- cos(500π) = 1
- sin(500π) = 0
Therefore, (1 + i)^2000 = 2^1000 (1 + 0i) = 2^1000
Conclusion: A Surprising Outcome
Despite the seemingly complex exponent, (1 + i)^2000 simplifies to a surprisingly simple real number: 2^1000. This demonstrates the power of De Moivre's Theorem in handling complex number operations, providing a concise and elegant solution.