%5E2023.jpeg "(1+i)^2023")
Exploring the Power of Complex Numbers: (1+i)^2023
The expression (1+i)^2023 might seem daunting at first glance. However, by utilizing the properties of complex numbers, we can break down this problem into manageable steps and arrive at a surprisingly elegant solution.
Understanding Complex Numbers
Before diving into the calculation, let's understand the basics of complex numbers. A complex number is a number that can be expressed in the form a + bi, where a and b are real numbers and i is the imaginary unit, defined as the square root of -1.
De Moivre's Theorem
The key to solving our problem lies in 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, the following equation holds:
(r(cos θ + i sin θ))^n = r^n(cos nθ + i sin nθ)
Applying De Moivre's Theorem
Let's rewrite (1+i) in polar form. Its magnitude is √2 and its argument is π/4. Therefore,
(1+i) = √2 (cos π/4 + i sin π/4)
Now, we can apply De Moivre's Theorem:
(1+i)^2023 = (√2 (cos π/4 + i sin π/4))^2023
= (√2)^2023 (cos (2023π/4) + i sin (2023π/4))
= 2^1011.5 (cos (505.75π) + i sin (505.75π))
Finding the Solution
Notice that 505.75π can be simplified. Since the cosine and sine functions have a period of 2π, we can subtract multiples of 2π from the argument without changing the result.
505.75π - 252 * 2π = 1.75π
Therefore:
(1+i)^2023 = 2^1011.5 (cos 1.75π + i sin 1.75π)
= 2^1011.5 (-1 + 0i)
= -2^1011.5
Conclusion
By utilizing De Moivre's Theorem and simplifying the resulting complex number, we have found that (1+i)^2023 = -2^1011.5. This demonstrates the power of complex numbers and how understanding their properties can lead to elegant solutions for seemingly complex problems.