(1+i)^2011/(1-i)^2009

Jasonbradley

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Jun 16, 2024

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Simplifying (1+i)^2011 / (1-i)^2009

This problem involves complex numbers and their powers. To simplify the expression (1+i)^2011 / (1-i)^2009, we can utilize the following steps:

1. Expressing in Polar Form:

• Finding the Magnitude:

  • |1+i| = √(1² + 1²) = √2
  • |1-i| = √(1² + (-1)²) = √2

• Finding the Argument:

  • Arg(1+i) = arctan(1/1) = π/4
  • Arg(1-i) = arctan(-1/1) = -π/4 (or 7π/4)

• Polar Form:

  • 1+i = √2 * cis(π/4)
  • 1-i = √2 * cis(-π/4)

2. Applying De Moivre's Theorem:

De Moivre's Theorem states: (cis θ)^n = cis(nθ).

Applying this to our problem:

• (1+i)^2011 = (√2 * cis(π/4))^2011 = 2^1005.5 * cis(2011π/4)
• (1-i)^2009 = (√2 * cis(-π/4))^2009 = 2^1004.5 * cis(-2009π/4)

3. Simplifying the Expression:

Now we can substitute the polar forms back into the original expression:

(1+i)^2011 / (1-i)^2009 = (2^1005.5 * cis(2011π/4)) / (2^1004.5 * cis(-2009π/4))

• Simplifying the Magnitude: 2^1005.5 / 2^1004.5 = 2
• Simplifying the Argument: cis(2011π/4) / cis(-2009π/4) = cis(2011π/4 + 2009π/4) = cis(2010π/2) = cis(1005π) = cis(π)

4. Final Result:

Therefore, (1+i)^2011 / (1-i)^2009 simplifies to 2 * cis(π) = -2.