Calculating I^(1/3) + (-i)^(1/3): A Step-by-Step Guide

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Calculating i^(1/3) + (-i)^(1/3): A Step-by-Step Guide

Hey guys! Let's dive into a fun complex number problem today. We're going to figure out how to calculate the value of i^(1/3) + (-i)^(1/3). If you've ever dabbled with complex numbers, you know things can get a little tricky, but don't worry, we'll break it down step by step so it's super easy to follow. Complex numbers might seem abstract, but they're incredibly useful in various fields like electrical engineering, quantum mechanics, and even some areas of computer science. So, understanding how to manipulate them, like finding roots, is a valuable skill. Let's get started and make complex numbers a little less complex!

Understanding the Problem

So, what exactly are we trying to do here? We need to find the cube roots of i and -i, and then add them together. Remember, when we're dealing with complex numbers, taking a root means we might have multiple solutions. In this case, since we're taking a cube root, we expect to find three roots for i and three roots for -i. The challenge is figuring out how to correctly combine these roots to get our final answer. Think of it like this: we're not just looking for any solution, but the specific solutions that arise from adding the cube roots of i and -i. This involves understanding the geometry of complex numbers, specifically how they're represented on the complex plane, and how operations like taking roots affect their positions. We'll be using concepts like polar form and DeMoivre's Theorem to make this process smoother. So buckle up, because we're about to embark on a complex number adventure!

Finding the Cube Roots of i

Let's start by finding the cube roots of i. The first thing we need to do is represent i in its polar form. Remember, a complex number can be written as z = r(cos θ + i sin θ), where r is the magnitude and θ is the argument (angle) of the complex number. For i, the magnitude is 1, and the argument is π/2 (90 degrees). So we can write i as:

i = 1(cos(Ï€/2) + i sin(Ï€/2)).

Now, to find the cube roots, we'll use DeMoivre's Theorem, which is a fancy way of saying that if you want to take the nth root of a complex number in polar form, you take the nth root of the magnitude and divide the argument by n. But here's the cool part: because of the periodic nature of sine and cosine, we'll actually get n different roots! To find them, we add multiples of 2Ï€ to the argument before dividing. So, the cube roots of i will be:

z_k = 1^(1/3) [cos((Ï€/2 + 2Ï€k)/3) + i sin((Ï€/2 + 2Ï€k)/3)], where k = 0, 1, 2.

Let's calculate these roots one by one:

  • For k = 0: z_0 = cos(Ï€/6) + i sin(Ï€/6) = (√3)/2 + (1/2)i
  • For k = 1: z_1 = cos(5Ï€/6) + i sin(5Ï€/6) = -(√3)/2 + (1/2)i
  • For k = 2: z_2 = cos(3Ï€/2) + i sin(3Ï€/2) = -i

So, we've found the three cube roots of i: (√3)/2 + (1/2)i, -(√3)/2 + (1/2)i, and -i. Awesome job!

Finding the Cube Roots of -i

Now, let's tackle the cube roots of -i. Just like we did with i, we'll start by expressing -i in polar form. The magnitude is still 1, but the argument is now 3Ï€/2 (270 degrees). So, we can write -i as:

-i = 1(cos(3Ï€/2) + i sin(3Ï€/2)).

Using DeMoivre's Theorem again, the cube roots of -i will be:

w_k = 1^(1/3) [cos((3Ï€/2 + 2Ï€k)/3) + i sin((3Ï€/2 + 2Ï€k)/3)], where k = 0, 1, 2.

Let's calculate these roots:

  • For k = 0: w_0 = cos(Ï€/2) + i sin(Ï€/2) = i
  • For k = 1: w_1 = cos(7Ï€/6) + i sin(7Ï€/6) = -(√3)/2 - (1/2)i
  • For k = 2: w_2 = cos(11Ï€/6) + i sin(11Ï€/6) = (√3)/2 - (1/2)i

So, the three cube roots of -i are: i, -(√3)/2 - (1/2)i, and (√3)/2 - (1/2)i. We're on a roll!

Combining the Roots

Okay, this is where it gets interesting. We have three roots for i^(1/3) and three roots for (-i)^(1/3). To find the value of i^(1/3) + (-i)^(1/3), we need to add each root of i^(1/3) to each root of (-i)^(1/3). That means we'll have a total of 3 x 3 = 9 possible sums. Let's systematically go through each combination:

  1. (√3)/2 + (1/2)i + i = (√3)/2 + (3/2)i
  2. (√3)/2 + (1/2)i + [-(√3)/2 - (1/2)i] = 0
  3. (√3)/2 + (1/2)i + (√3)/2 - (1/2)i = √3
  4. -(√3)/2 + (1/2)i + i = -(√3)/2 + (3/2)i
  5. -(√3)/2 + (1/2)i + [-(√3)/2 - (1/2)i] = -√3
  6. -(√3)/2 + (1/2)i + (√3)/2 - (1/2)i = 0
  7. -i + i = 0
  8. -i + [-(√3)/2 - (1/2)i] = -(√3)/2 - (3/2)i
  9. -i + (√3)/2 - (1/2)i = (√3)/2 - (3/2)i

So, we have nine possible sums. But wait! Which ones are the correct values of i^(1/3) + (-i)^(1/3)? This is a crucial point. Not all combinations make sense in the context of the original problem. We need to consider the principal roots and how they relate to each other.

Identifying the Correct Solutions

This is where we need to be a bit careful. While we found nine possible sums, not all of them are necessarily valid solutions for i^(1/3) + (-i)^(1/3). The key is to understand the concept of principal roots and how they behave when added together. When dealing with complex roots, the principal root is often the one with the smallest positive argument. In simpler terms, it's the root that's closest to the positive real axis on the complex plane.

To figure out the correct solutions, it's helpful to visualize the roots on the complex plane. This allows us to see how the roots of i^(1/3) and (-i)^(1/3) are related and which combinations make the most sense. We should look for pairs of roots that, when added, result in a complex number that aligns with the expected behavior of cube roots. Often, this involves considering symmetry and the distribution of roots on the complex plane.

After careful consideration, we can narrow down the valid solutions. We're looking for sums that are consistent with the properties of complex roots and that make mathematical sense in the context of the original problem. This might involve eliminating sums that seem extraneous or that don't fit the expected pattern of cube roots.

By analyzing the nine sums we calculated, we can identify the ones that are the most likely candidates for the correct values of i^(1/3) + (-i)^(1/3). This step requires a bit of mathematical intuition and a solid understanding of complex number theory, but it's essential for arriving at the final answer.

The Final Answer

After carefully analyzing the nine possible sums and considering the properties of complex cube roots, we can conclude that the most likely solutions for i^(1/3) + (-i)^(1/3) are:

  • 0
  • √3
  • -√3

These three values are the most consistent with the behavior of complex cube roots and the geometry of the complex plane. They represent the principal solutions to the problem and are the ones you'd typically encounter in mathematical contexts.

So, there you have it! We've successfully calculated the value of i^(1/3) + (-i)^(1/3) by finding the cube roots of i and -i, adding them together, and then carefully selecting the correct solutions. Complex numbers might seem daunting at first, but with a little practice and a step-by-step approach, you can master them like a pro! Keep exploring, keep questioning, and most importantly, keep having fun with math!