Cube Root Of -8-8i Location On The Unit Circle

Hey guys! Today, we're diving into a fascinating math problem that involves complex numbers and the unit circle. Specifically, we're tackling the question: Which region on the unit circle contains a cube root of -8-8i? This isn't just a theoretical exercise; understanding complex roots is crucial in various fields, from electrical engineering to quantum mechanics. So, let's break it down step by step, making sure everyone, even those who aren't math whizzes, can follow along.

Understanding the Problem: Cube Roots of Complex Numbers

Before we jump into solving the problem, let's make sure we're all on the same page about the basics. What exactly are we looking for when we talk about the cube root of a complex number like -8-8i? A cube root of a number, whether it's real or complex, is a value that, when multiplied by itself three times, gives you the original number. In mathematical terms, if z is a cube root of -8-8i, then z³ = -8-8i. Now, complex numbers, unlike real numbers, can have multiple roots. In fact, a complex number generally has n distinct n-th roots. So, a complex number like -8-8i will have three distinct cube roots.

Why three roots? This stems from the fundamental theorem of algebra, which states that a polynomial equation of degree n has n complex roots (counting multiplicities). The equation z³ = -8-8i is a polynomial equation of degree 3, hence three roots. These roots aren't just random numbers; they have a beautiful geometric interpretation on the complex plane, which is where the unit circle comes into play. The unit circle is a circle with a radius of 1 centered at the origin of the complex plane. Complex numbers can be represented as points on this plane, and their roots are evenly spaced around the circle. This is a key concept for visualizing and understanding the solutions to our problem.

To find these cube roots, we'll need to represent -8-8i in its polar form. The polar form of a complex number expresses it in terms of its magnitude (distance from the origin) and its argument (angle from the positive real axis). This representation makes it much easier to calculate roots. Think of it like switching from Cartesian coordinates (x, y) to polar coordinates (r, θ) – sometimes one system is more convenient than the other, and in this case, polar form is our friend.

So, with these foundational concepts in mind, we're ready to dive into the specific steps of finding the cube roots of -8-8i and determining where they lie on the unit circle. It might seem daunting now, but we'll break it down into manageable pieces, ensuring you grasp the underlying principles every step of the way. Stay tuned as we transform this complex problem into a clear and understandable solution!

Step-by-Step Solution: Finding the Cube Roots

Alright, let's get down to business and solve this problem step-by-step. Our main goal is to find the cube roots of the complex number -8-8i and then pinpoint which region of the unit circle these roots fall into. To do this, we'll follow a structured approach:

1. Convert to Polar Form: The first crucial step is to convert the complex number -8-8i into its polar form. Remember, the polar form of a complex number is expressed as r(cos θ + i sin θ), where r is the magnitude and θ is the argument. The magnitude r is calculated as the square root of the sum of the squares of the real and imaginary parts. In our case, r = √((-8)² + (-8)²) = √(128) = 8√2. Now, to find the argument θ, we use the arctangent function: θ = arctan(imaginary part / real part) = arctan(-8 / -8) = arctan(1). Since -8-8i lies in the third quadrant (both real and imaginary parts are negative), we need to add π to the arctan result to get the correct angle. Thus, θ = π/4 + π = 5π/4. So, -8-8i in polar form is 8√2 (cos(5π/4) + i sin(5π/4)). This polar form is our gateway to easily finding the cube roots.

2. Apply De Moivre's Theorem: This is where De Moivre's Theorem comes to our rescue. This theorem provides a neat formula for finding the n-th roots of a complex number in polar form. It states that if z = r(cos θ + i sin θ), then the n-th roots of z are given by zₖ = r^(1/n) [cos((θ + 2πk)/n) + i sin((θ + 2πk)/n)] for k = 0, 1, 2, ..., n-1. In our case, we want the cube roots (n = 3) of 8√2 (cos(5π/4) + i sin(5π/4)). So, we'll apply De Moivre's Theorem with n = 3 and k ranging from 0 to 2.

3. Calculate the Roots: Now, we plug in the values and calculate the three cube roots. Let's do it for each value of k:

   • For k = 0: z₀ = (8√2)^(1/3) [cos((5π/4 + 2π0)/3) + i sin((5π/4 + 2π0)/3)] ≈ 2 [cos(5π/12) + i sin(5π/12)]
   • For k = 1: z₁ = (8√2)^(1/3) [cos((5π/4 + 2π1)/3) + i sin((5π/4 + 2π1)/3)] ≈ 2 [cos(13π/12) + i sin(13π/12)]
   • For k = 2: z₂ = (8√2)^(1/3) [cos((5π/4 + 2π2)/3) + i sin((5π/4 + 2π2)/3)] ≈ 2 [cos(21π/12) + i sin(21π/12)] = 2 [cos(7π/4) + i sin(7π/4)]

4. Determine the Quadrant: The final step is to figure out which quadrants these roots lie in. Remember, the unit circle is divided into four quadrants:

   • Quadrant I: 0 < θ < π/2
   • Quadrant II: π/2 < θ < π
   • Quadrant III: π < θ < 3π/2
   • Quadrant IV: 3π/2 < θ < 2π

   Let's analyze each root:

   • z₀ has an argument of 5π/12, which falls between 0 and π/2, placing it in Quadrant I.
   • z₁ has an argument of 13π/12, which falls between π and 3π/2, placing it in Quadrant III.
   • z₂ has an argument of 7π/4, which falls between 3π/2 and 2π, placing it in Quadrant IV.

So, we've found that one of the cube roots lies in Quadrant I, one in Quadrant III, and one in Quadrant IV. Now, let's circle back to the original question and choose the correct answer.

Final Answer: Pinpointing the Correct Region

Okay, we've done the heavy lifting! We converted -8-8i to polar form, applied De Moivre's Theorem, calculated the three cube roots, and determined the quadrant each root lies in. Now, let's revisit the original question: Which region on the unit circle contains a cube root of -8-8i?

We found that the cube roots lie in Quadrant I, Quadrant III, and Quadrant IV. Looking at the answer choices, we have:

A. in quadrant I B. in quadrant II C. on the imaginary axis D. on the real axis

Clearly, the correct answer is A. in quadrant I. We identified that the cube root z₀ with an argument of 5π/12 falls squarely within the boundaries of Quadrant I. While the other cube roots are important and part of the complete solution, the question specifically asks for a cube root, and we've confidently located one in Quadrant I.

This problem beautifully illustrates how complex numbers, polar form, De Moivre's Theorem, and the unit circle all come together. It's not just about memorizing formulas; it's about understanding the relationships between these concepts and how they help us solve problems. By breaking down the problem into smaller, manageable steps, we've navigated the complexities and arrived at a clear and accurate solution. Remember, guys, math is like a puzzle, and every piece has its place. The more you practice, the better you'll become at fitting those pieces together!

Why This Matters: Applications of Complex Roots

So, we've successfully found the cube root of -8-8i that lies in Quadrant I. But you might be wondering,