%5E200.jpeg "(1+i)^200")
DeMoivre's Theorem and the Power of Complex Numbers: Exploring (1+i)^200
The expression (1+i)^200 might seem daunting at first glance, but with the help of DeMoivre's Theorem, we can tackle this calculation efficiently. DeMoivre's Theorem provides a powerful tool for working with complex numbers raised to powers.
Understanding DeMoivre's Theorem
DeMoivre's Theorem states:
For any complex number z = r(cos θ + i sin θ) and any integer n, z^n = r^n(cos nθ + i sin nθ).
In simpler terms, this means:
- Raise the modulus (r) to the power (n).
- Multiply the argument (θ) by the power (n).
Applying DeMoivre's Theorem to (1+i)^200
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Convert (1+i) to polar form:
- Modulus (r): |1+i| = √(1^2 + 1^2) = √2
- Argument (θ): θ = tan⁻¹(1/1) = π/4 (since 1+i lies in the first quadrant)
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Apply DeMoivre's Theorem: (1+i)^200 = (√2)^200 * (cos(200 * π/4) + i sin(200 * π/4))
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Simplify: (1+i)^200 = 2^100 * (cos(50π) + i sin(50π))
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Reduce the argument:
- Since cos(2πn) = 1 and sin(2πn) = 0 for any integer n: (1+i)^200 = 2^100 * (1 + 0i) = 2^100
Conclusion
Therefore, (1+i)^200 simplifies to the remarkably simple expression 2^100. This exemplifies the power of DeMoivre's Theorem in dealing with complex number powers, allowing us to find elegant solutions for seemingly complex calculations.