(2-2i)^8

3 min read Jun 16, 2024
(2-2i)^8

Exploring the Power of Complex Numbers: (2-2i)^8

This article delves into the process of calculating the complex number (2-2i)^8. We'll utilize De Moivre's Theorem, a powerful tool for simplifying complex number exponents.

Understanding De Moivre's Theorem

De Moivre's Theorem states that for any complex number in polar form, z = r(cos θ + i sin θ) and any integer n:

z^n = r^n (cos(nθ) + i sin(nθ))

This theorem allows us to calculate powers of complex numbers efficiently.

Converting to Polar Form

Firstly, we need to express (2-2i) in polar form. We can achieve this by finding the modulus (r) and argument (θ) of the complex number.

  • Modulus (r): |2-2i| = √(2² + (-2)²) = √8 = 2√2
  • Argument (θ): θ = arctan(-2/2) = -π/4. (We choose -π/4 as the complex number lies in the fourth quadrant)

Therefore, (2-2i) in polar form is 2√2 (cos(-π/4) + i sin(-π/4))

Applying De Moivre's Theorem

Now, we can apply De Moivre's Theorem to calculate (2-2i)^8:

(2-2i)^8 = (2√2)^8 * (cos(-8π/4) + i sin(-8π/4))

Simplifying:

(2-2i)^8 = 256 * (cos(-2π) + i sin(-2π))

Since cos(-2π) = 1 and sin(-2π) = 0, the result is:

(2-2i)^8 = 256

Conclusion

By utilizing De Moivre's Theorem, we successfully calculated (2-2i)^8 and obtained the surprising result of 256. This demonstrates the power of De Moivre's Theorem in simplifying complex number calculations and highlights the fascinating properties of complex numbers.

Related Post


Featured Posts