(1+i)^n+(1-i)^n =2^n+2/2 Cos (nπ/4)

Jasonbradley

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Jun 16, 2024

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Exploring the Relationship: (1+i)^n + (1-i)^n = 2^(n/2) * cos(nπ/4)

This article delves into the intriguing relationship between complex numbers and trigonometric functions, exploring the equation:

(1 + i)^n + (1 - i)^n = 2^(n/2) * cos(nπ/4)

Let's break down this equation and understand why it holds true.

Understanding the Basics

• Complex Numbers: A complex number is a number of the form a + bi, where 'a' and 'b' are real numbers, and 'i' is the imaginary unit (√-1).
• De Moivre's Theorem: This theorem states that for any complex number in polar form (r(cos θ + i sin θ)) and any integer n, (r(cos θ + i sin θ))^n = r^n(cos nθ + i sin nθ).

Proving the Equation

1. Expressing in Polar Form:

   • (1 + i) can be written in polar form as √2(cos(π/4) + i sin(π/4))
   • (1 - i) can be written in polar form as √2(cos(-π/4) + i sin(-π/4))

2. Applying De Moivre's Theorem:

   • (1 + i)^n = (√2)^n (cos(nπ/4) + i sin(nπ/4))
   • (1 - i)^n = (√2)^n (cos(-nπ/4) + i sin(-nπ/4))

3. Simplifying:

   • Since cos(-θ) = cos(θ) and sin(-θ) = -sin(θ), we can rewrite (1 - i)^n as:
   • (1 - i)^n = (√2)^n (cos(nπ/4) - i sin(nπ/4))

4. Adding the Terms:

   • (1 + i)^n + (1 - i)^n = (√2)^n(cos(nπ/4) + i sin(nπ/4)) + (√2)^n(cos(nπ/4) - i sin(nπ/4))
   • This simplifies to 2^(n/2) * 2 * cos(nπ/4) = 2^(n/2) * cos(nπ/4)

Conclusion

We have successfully demonstrated that the equation (1 + i)^n + (1 - i)^n = 2^(n/2) * cos(nπ/4) is true. This equation showcases the elegant relationship between complex numbers and trigonometric functions. De Moivre's theorem proves to be a valuable tool for simplifying and understanding such complex expressions.

This exploration can serve as a foundation for further exploration into the fascinating world of complex numbers and their diverse applications across various fields of mathematics and science.