%5E4m%3D1.jpeg "(1+i/1-i)^4m=1")
Exploring the Complex Equation: (1+i/1-i)^4m = 1
This article will delve into the intriguing equation (1+i/1-i)^4m = 1, where 'i' represents the imaginary unit (√-1) and 'm' is an integer. We'll explore the properties of complex numbers and utilize De Moivre's Theorem to understand the solution.
Simplifying the Expression
First, let's simplify the expression within the parentheses:
- (1+i)/(1-i) can be simplified by multiplying both numerator and denominator by the conjugate of the denominator (1+i):
[(1+i)(1+i)]/[(1-i)(1+i)] = (1 + 2i + i^2) / (1 - i^2)
- Since i² = -1, we get:
(1 + 2i - 1) / (1 + 1) = 2i / 2 = i
Therefore, our equation now becomes: i^4m = 1.
Utilizing De Moivre's Theorem
De Moivre's 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)
Let's express 'i' in polar form:
- Magnitude: |i| = 1
- Angle: The angle of 'i' is 90 degrees or π/2 radians.
Therefore, i can be written as: 1(cos π/2 + i sin π/2).
Applying De Moivre's Theorem to our simplified equation:
(1(cos π/2 + i sin π/2))^4m = 1(cos 4mπ/2 + i sin 4mπ/2)
For this equation to hold true, the cosine term must equal 1 and the sine term must equal 0. This occurs when 4mπ/2 is a multiple of 2π.
4mπ/2 = 2πk (where 'k' is an integer)
Simplifying the equation:
2mπ = 2πk
m = k
Therefore, the equation (1+i/1-i)^4m = 1 is true for all integer values of 'm'.
Conclusion
The equation (1+i/1-i)^4m = 1 demonstrates the fascinating properties of complex numbers. By simplifying the expression, applying De Moivre's Theorem, and analyzing the conditions for the equation to hold true, we found that the solution is valid for all integer values of 'm'. This highlights the cyclical nature of complex numbers and their ability to generate solutions through elegant mathematical techniques.