%5En%2B(1-i)%5En%3D2%5E(n%2F2%2B1)x-cos(pi)%2F4.jpeg "(1+i)^n+(1-i)^n=2^(n/2+1)x Cos(pi)/4")
DeMoivre's Theorem and the Proof of (1+i)^n + (1-i)^n = 2^(n/2+1) * cos(π/4)
This article explores the fascinating relationship between complex numbers, DeMoivre's Theorem, and the trigonometric function cosine. Specifically, we will prove the equation:
(1 + i)^n + (1 - i)^n = 2^(n/2 + 1) * cos(π/4)
where n is a positive integer.
Understanding the Concepts
- Complex Numbers: Complex numbers are of the form a + bi, where a and b are real numbers, and i is the imaginary unit, satisfying i² = -1.
- DeMoivre's Theorem: This powerful theorem states that for any complex number z = r(cos θ + i sin θ) and any integer n, the following holds:
- z^n = r^n(cos(nθ) + i sin(nθ))
- Trigonometric Functions: The cosine function (cos) is a periodic function that maps angles to real numbers, with values ranging from -1 to 1.
Proof of the Equation
-
Expressing (1+i) and (1-i) in Polar Form:
- (1+i) has a magnitude (modulus) of √2 and an angle (argument) of π/4. Therefore, in polar form:
- (1 + i) = √2(cos(π/4) + i sin(π/4))
- (1-i) has the same magnitude as (1+i) but an angle of -π/4. In polar form:
- (1 - i) = √2(cos(-π/4) + i sin(-π/4))
- (1+i) has a magnitude (modulus) of √2 and an angle (argument) of π/4. Therefore, in polar form:
-
Applying DeMoivre's Theorem:
- Using DeMoivre's Theorem, we can raise both (1+i) and (1-i) to the power of n:
- (1 + i)^n = (√2)^n(cos(nπ/4) + i sin(nπ/4))
- (1 - i)^n = (√2)^n(cos(-nπ/4) + i sin(-nπ/4))
- Using DeMoivre's Theorem, we can raise both (1+i) and (1-i) to the power of n:
-
Simplifying and Combining:
- Notice that cos(-x) = cos(x) and sin(-x) = -sin(x). Therefore:
- (1 - i)^n = (√2)^n(cos(nπ/4) - i sin(nπ/4))
- Now, adding (1+i)^n and (1-i)^n, the imaginary terms cancel out:
- (1 + i)^n + (1 - i)^n = (√2)^n(2cos(nπ/4))
- Notice that cos(-x) = cos(x) and sin(-x) = -sin(x). Therefore:
-
Final Result:
- Simplifying further:
- (√2)^n = 2^(n/2)
- (1 + i)^n + (1 - i)^n = 2^(n/2) * 2 * cos(nπ/4)
- (1 + i)^n + (1 - i)^n = 2^(n/2 + 1) * cos(nπ/4)
- Simplifying further:
Conclusion
Through the application of DeMoivre's Theorem and the simplification of complex numbers in polar form, we have successfully proved that (1 + i)^n + (1 - i)^n = 2^(n/2 + 1) * cos(π/4) for any positive integer n. This equation reveals an elegant connection between complex exponentiation, trigonometric functions, and the power of mathematical tools like DeMoivre's Theorem.