(1+i)^2n=(1-i)^2n

Jasonbradley

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Jun 16, 2024

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Solving the Equation: (1+i)^2n = (1-i)^2n

This article explores the solution to the equation (1+i)^2n = (1-i)^2n, where 'i' represents the imaginary unit (√-1) and 'n' is an integer.

Understanding Complex Numbers

Before delving into the solution, let's recall some key properties of complex numbers:

• Polar Form: Complex numbers can be represented in polar form as r(cosθ + isinθ), where 'r' is the magnitude and 'θ' is the angle.
• De Moivre's Theorem: This theorem states that (cosθ + isinθ)^n = cos(nθ) + isin(nθ).

Solving the Equation

1. Expressing in Polar Form:

   • (1+i) = √2(cos(π/4) + isin(π/4))
   • (1-i) = √2(cos(-π/4) + isin(-π/4))

2. Applying De Moivre's Theorem:

   • (1+i)^2n = (√2)^2n * (cos(nπ/2) + isin(nπ/2))
   • (1-i)^2n = (√2)^2n * (cos(-nπ/2) + isin(-nπ/2))

3. Equating:

   • (√2)^2n * (cos(nπ/2) + isin(nπ/2)) = (√2)^2n * (cos(-nπ/2) + isin(-nπ/2))

4. Simplifying:

   • cos(nπ/2) + isin(nπ/2) = cos(-nπ/2) + isin(-nπ/2)

5. Finding Solutions:

   • For the equation to hold true, the real and imaginary components must be equal. This means:
     • cos(nπ/2) = cos(-nπ/2)
     • sin(nπ/2) = sin(-nπ/2)
   • These conditions are satisfied when n is an even integer.

Conclusion

Therefore, the equation (1+i)^2n = (1-i)^2n is true for all even integer values of 'n'. This solution arises from the properties of complex numbers, particularly De Moivre's Theorem, which allows us to simplify and analyze the equation effectively.