Exploring the Power of Complex Numbers: (1 + i)^40
In the realm of complex numbers, exploring the powers of a complex number can lead to fascinating results. One such example is calculating (1 + i)^40. This might seem daunting at first, but with the right approach, it becomes a manageable problem.
Understanding Complex Numbers and De Moivre's Theorem
Complex numbers are expressed in the form a + bi, where 'a' and 'b' are real numbers and 'i' is the imaginary unit (√-1). To handle powers of complex numbers, we use 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:
(r(cos θ + i sin θ))^n = r^n(cos nθ + i sin nθ)
Converting to Polar Form
To apply De Moivre's Theorem, we need to express (1 + i) in polar form.
- Magnitude: The magnitude, or modulus, of a complex number is its distance from the origin in the complex plane. For (1 + i), this is √(1^2 + 1^2) = √2.
- Angle: The angle, or argument, is the angle the line connecting the origin and the complex number makes with the positive real axis. For (1 + i), this angle is 45° or π/4 radians (found using the arctangent function).
Therefore, (1 + i) in polar form is √2 (cos π/4 + i sin π/4).
Applying De Moivre's Theorem
Now, we can use De Moivre's Theorem to calculate (1 + i)^40:
(√2 (cos π/4 + i sin π/4))^40 = (√2)^40 (cos (40 * π/4) + i sin (40 * π/4))
Simplifying:
(√2)^40 (cos 10π + i sin 10π) = 2^20 (cos 0 + i sin 0)
Since cos 0 = 1 and sin 0 = 0:
(1 + i)^40 = 2^20 = 1,048,576
Conclusion
By using De Moivre's Theorem and converting to polar form, we find that (1 + i)^40 = 1,048,576. This demonstrates the power of complex numbers and the elegant tools we have to manipulate them. Despite the seeming complexity of the problem, with the right methods, we can arrive at a relatively simple solution.