%5E5.jpeg "(2-2i)^5")
Exploring the Power of Complex Numbers: (2 - 2i)^5
This article explores the expansion of the complex number (2 - 2i) raised to the power of 5. We'll delve into the process of calculating this, utilizing the concept of De Moivre's Theorem, and ultimately unveil the fascinating result.
Understanding the Basics
Before tackling the exponent, let's understand the basics of complex numbers:
- Complex numbers are numbers that consist of a real part and an imaginary part. They are typically written in the form a + bi, where a and b are real numbers, and i is the imaginary unit (√-1).
- De Moivre's Theorem provides a powerful tool to calculate powers of complex numbers. It states that for any complex number in polar form, r(cos θ + i sin θ), its power n can be calculated as:
[r(cos θ + i sin θ)]^n = r^n (cos nθ + i sin nθ)
Converting to Polar Form
First, we need to express (2 - 2i) in polar form. We can visualize this number as a point in the complex plane:
- Magnitude (r): The magnitude of (2 - 2i) is calculated as √(2^2 + (-2)^2) = √8 = 2√2.
- Angle (θ): The angle θ is found by considering the arctangent of the imaginary part over the real part, adjusted for the correct quadrant: θ = arctan(-2/2) = -π/4 (since the point 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)^5:
(2√2 (cos (-π/4) + i sin (-π/4)))^5 = (2√2)^5 (cos (-5π/4) + i sin (-5π/4))
This simplifies to:
128 (cos (-5π/4) + i sin (-5π/4))
Converting Back to Rectangular Form
Finally, we can convert this back to rectangular form:
128 (cos (-5π/4) + i sin (-5π/4)) = 128 (-√2/2 + i√2/2) = -64√2 + 64√2 i
Conclusion
Therefore, (2 - 2i)^5 is equal to -64√2 + 64√2 i. This exercise demonstrates the power and elegance of De Moivre's Theorem in simplifying complex calculations involving powers of complex numbers.