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

3 min read Jun 16, 2024

Exploring the Pattern: (1+i)²ⁿ + (1-i)²ⁿ

This article delves into the intriguing expression (1+i)²ⁿ + (1-i)²ⁿ, where 'i' represents the imaginary unit (√-1) and 'n' is any positive integer. We'll explore its fascinating properties and uncover a hidden pattern that simplifies its calculation.

Demystifying the Expression

At first glance, the expression might seem complex. However, by expanding and simplifying, we can reveal its underlying structure. Let's start by examining the individual terms:

(1+i)²ⁿ: Expanding this using the binomial theorem, we get a series of terms involving powers of 'i'. Remember that i² = -1, i³ = -i, and i⁴ = 1.
(1-i)²ⁿ: Similarly, expanding this term will also yield a series of terms involving powers of 'i'.

When we add these two expanded expressions together, we observe a remarkable cancellation. The terms involving odd powers of 'i' will cancel out due to their opposite signs. This leaves us with only the terms containing even powers of 'i' (i², i⁴, i⁶, etc.).

Unveiling the Pattern

Since i² = -1, all even powers of 'i' reduce to either +1 or -1. This leads to a simplified pattern:

If 'n' is even: All the even powers of 'i' will result in +1. Therefore, (1+i)²ⁿ + (1-i)²ⁿ will be a sum of positive terms, resulting in a positive value.
If 'n' is odd: The even powers of 'i' will alternate between +1 and -1. This creates a cancellation effect, resulting in a zero value.

The Final Result

Based on this pattern, we can express the value of (1+i)²ⁿ + (1-i)²ⁿ concisely:

(1+i)²ⁿ + (1-i)²ⁿ = 2ⁿ if 'n' is even
(1+i)²ⁿ + (1-i)²ⁿ = 0 if 'n' is odd

Conclusion

The seemingly complex expression (1+i)²ⁿ + (1-i)²ⁿ simplifies to a neat and elegant pattern. By understanding the properties of the imaginary unit and applying the binomial theorem, we uncover a hidden relationship between the even and odd values of 'n'. This exploration demonstrates the beauty and elegance of mathematical patterns and how simplification can often unveil deeper insights.