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

2 min read Jun 16, 2024

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:

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.