%5E1000.jpeg "(1+i)^1000")
Exploring the Complex Power: (1 + i)^1000
The expression (1 + i)^1000 might seem daunting at first glance, especially considering the complex number 'i' and the massive exponent. However, with the right tools from complex number theory, we can elegantly tackle this problem.
De Moivre's Theorem: A Powerful Tool
The key to simplifying this expression lies in De Moivre's Theorem. This theorem states that for any complex number in polar form, z = r(cos θ + i sin θ), and any integer n, the following holds true:
z^n = r^n (cos (nθ) + i sin (nθ))
Transforming (1 + i) into Polar Form
To apply De Moivre's Theorem, we need to express (1 + i) in polar form.
- Magnitude: The magnitude of (1 + i) is √(1² + 1²) = √2.
- Angle: The angle θ can be found using arctan(1/1) = π/4 radians.
Therefore, (1 + i) can be written as √2(cos(π/4) + i sin(π/4)).
Applying De Moivre's Theorem
Now we can apply De Moivre's Theorem to (1 + i)^1000:
(1 + i)^1000 = (√2(cos(π/4) + i sin(π/4)))^1000
This simplifies to:
(√2)^1000 (cos(1000π/4) + i sin(1000π/4))
Simplifying the Result
- Magnitude: (√2)^1000 = 2^500
- Angle: 1000π/4 = 250π. Since the cosine and sine functions have a period of 2π, we can simplify this to cos(0) + i sin(0).
Therefore, the final simplified form of (1 + i)^1000 is:
(1 + i)^1000 = 2^500 (cos(0) + i sin(0)) = 2^500
Conclusion
Through the application of De Moivre's Theorem, we have elegantly simplified (1 + i)^1000 to the real number 2^500. This demonstrates the power of complex number theory in solving seemingly complex problems.