(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.

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