Explanation:
Okay, so I need to find the smallest positive integer n such that all the roots of the equation z⁴ + z² + 1 = 0 are nth roots of unity. Hmm, let me start by recalling what roots of unity are. A root of unity is a complex number that satisfies zⁿ = 1 for some positive integer n. So, essentially, these are the solutions to the equation zⁿ = 1. The roots of unity are located on the unit circle in the complex plane and are equally spaced around the circle.

Now, the problem is about the roots of the polynomial z⁴ + z² + 1 = 0. I need to find the smallest n such that each of these roots is an nth root of unity. That means each root should satisfy zⁿ = 1. So, first, maybe I should find the roots of the given polynomial and then determine what n would make them roots of unity.

But solving z⁴ + z² + 1 = 0 directly might be a bit tricky. Let me think. Maybe I can factor this polynomial or relate it to a known equation. Let me try substituting w = z². Then the equation becomes w² + w + 1 = 0. That seems simpler. Let me solve for w first.

So, solving w² + w + 1 = 0. Using the quadratic formula: w = [-1 ± √(1 - 4)] / 2 = [-1 ± √(-3)] / 2 = (-1 ± i√3)/2. Therefore, the solutions for w are complex numbers. Then, since w = z², we can find z by taking square roots of these w's.

So, let me write down the roots. Let me denote the roots of w² + w + 1 = 0 as w₁ and w₂. Then, each root z of the original equation is a square root of w₁ or w₂. Let me compute those square roots.

First, let me recall that the roots of w² + w + 1 = 0 are the primitive third roots of unity. Wait, because the equation w² + w + 1 = 0 is the cyclotomic polynomial Φ₃(w), which factors as (w - e^(2πi/3))(w - e^(4πi/3)). So, these are the primitive 3rd roots of unity. Therefore, w₁ = e^(2πi/3) and w₂ = e^(4πi/3). Therefore, the original equation z⁴ + z² + 1 = 0 has roots which are square roots of these w's.

So, to find z, we need to take square roots of e^(2πi/3) and e^(4πi/3). Let me recall that taking square roots of a complex number on the unit circle involves taking the square root of the modulus (which is 1 here) and halving the angle. So, for example, the square roots of e^(2πi/3) would be e^(πi/3) and e^(πi/3 + πi) = e^(4πi/3). Wait, but hold on, let me check that.

Wait, if I have a complex number e^(iθ), then its square roots are e^(iθ/2) and e^(iθ/2 + πi). So, for θ = 2π/3, the square roots would be e^(iπ/3) and e^(iπ/3 + iπ) = e^(i4π/3). Similarly, for θ = 4π/3, the square roots would be e^(i2π/3) and e^(i2π/3 + iπ) = e^(i5π/3). Therefore, the roots of the original equation z⁴ + z² + 1 = 0 are e^(iπ/3), e^(i4π/3), e^(i2π/3), and e^(i5π/3). Wait, but let me check if that's correct.

Wait, let me verify. Let me compute z⁴ + z² + 1 for z = e^(iπ/3). Let me compute z² first. z² = e^(i2π/3). Then z⁴ = (z²)² = e^(i4π/3). Therefore, z⁴ + z² + 1 = e^(i4π/3) + e^(i2π/3) + 1. Let me compute this sum. Let me recall that e^(i2π/3) + e^(i4π/3) + 1 = 0, because they are the roots of w² + w + 1 = 0. Wait, but here we have e^(i4π/3) + e^(i2π/3) + 1 = 0. Therefore, z⁴ + z² + 1 = 0. So yes, that works. Similarly, for z = e^(i2π/3), z² = e^(i4π/3), z⁴ = e^(i8π/3) = e^(i(8π/3 - 2π)) = e^(i2π/3). Then z⁴ + z² + 1 = e^(i2π/3) + e^(i4π/3) + 1 = 0. So that works. Similarly for the other roots. Therefore, the roots of the original equation are e^(iπ/3), e^(i2π/3), e^(i4π/3), and e^(i5π/3). Wait, but e^(i5π/3) is the same as e^(-iπ/3), right? So, these are the 6th roots of unity? Wait, let me check.

Wait, e^(iπ/3) is a primitive 6th root of unity? Let me recall that the 6th roots of unity are e^(i2πk/6) for k = 0, 1, 2, 3, 4, 5. So, e^(iπ/3) is e^(i2π/6), which is a primitive 6th root of unity. Similarly, e^(i2π/3) is e^(i4π/6), which is a primitive 3rd root of unity. Wait, but 2π/3 is 4π/6. So, e^(i2π/3) is a primitive 3rd root of unity. Similarly, e^(i4π/3) is e^(i8π/6) which is e^(i4π/3), which is also a primitive 3rd root of unity. And e^(i5π/3) is e^(i10π/6) = e^(i5π/3), which is a primitive 6th root of unity. Therefore, the roots of the original equation are two primitive 3rd roots of unity and two primitive 6th roots of unity. Therefore, they are all roots of unity, but with different orders.

Therefore, to find the smallest n such that all these roots are nth roots of unity, we need to find the least common multiple (LCM) of the orders of these roots. The orders of the roots are 3 and 6. The LCM of 3 and 6 is 6. Therefore, n should be 6? Wait, but let me check if all roots are 6th roots of unity. Let me check each root:

- e^(iπ/3): This is e^(i2π/6), so it's a 6th root of unity. Similarly, e^(i5π/3) is also a 6th root of unity. e^(i2π/3) is e^(i4π/6), which is a 3rd root of unity, but since 3 divides 6, it is also a 6th root of unity. Similarly, e^(i4π/3) is e^(i8π/6) = e^(i4π/3), which is also a 6th root of unity. Therefore, all roots are 6th roots of unity. Therefore, n=6?

But wait, let me check if there is a smaller n. For example, n=3? Let me check if all roots are 3rd roots of unity. The roots e^(iπ/3) and e^(i5π/3) are not 3rd roots of unity because their angles are π/3 and 5π/3, which are not multiples of 2π/3. For example, e^(iπ/3) raised to the 3rd power is e^(iπ) = -1 ≠ 1. Therefore, they are not 3rd roots of unity. Therefore, n=3 is too small. Similarly, n=4? Let me check. The roots are e^(iπ/3), e^(i2π/3), e^(i4π/3), e^(i5π/3). Let me check if they are 4th roots of unity. The 4th roots of unity are e^(i0), e^(iπ/2), e^(iπ), e^(i3π/2). None of the roots we have are among these. For example, e^(iπ/3) is not a 4th root of unity. Therefore, n=4 is too small. Similarly, n=5? The 5th roots of unity are e^(i2πk/5) for k=0,...,4. Our roots have angles π/3, 2π/3, 4π/3, 5π/3. These angles are not multiples of 2π/5, so they are not 5th roots of unity. Therefore, n=5 is too small. Therefore, n=6 is the minimal?

Wait, but let me check if there is a smaller n than 6. For example, n=12? Wait, no, we need the minimal n. Wait, but maybe n=6 is the minimal. But let me think again. Let me check if all roots are 6th roots of unity. Since they are, and since there is no smaller n that works, then n=6 is the answer. But let me check if maybe there is a smaller n. For example, n=12? No, that's bigger. Wait, but maybe n= something else?

Alternatively, perhaps I can think algebraically. Let me note that the original equation z⁴ + z² + 1 = 0. Let me try to relate this to cyclotomic polynomials. Let me recall that z⁶ - 1 = (z³ - 1)(z³ + 1) = (z - 1)(z² + z + 1)(z + 1)(z² - z + 1). Also, z⁶ - 1 = (z² - 1)(z⁴ + z² + 1). Wait, let me check that. Let me compute (z² - 1)(z⁴ + z² + 1) = z⁶ + z⁴ + z² - z⁴ - z² - 1 = z⁶ - 1. Yes! Therefore, z⁴ + z² + 1 = (z⁶ - 1)/(z² - 1). Therefore, the roots of z⁴ + z² + 1 = 0 are the roots of z⁶ - 1 = 0 except for the roots of z² - 1 = 0. Therefore, the roots of z⁴ + z² + 1 = 0 are the 6th roots of unity except for z = 1 and z = -1. Therefore, the roots are e^(i2πk/6) for k = 1, 2, 3, 4, 5, excluding k=3 (which gives z=-1) and k=0 (which gives z=1). Wait, let me check:

Wait, z⁶ - 1 = 0 has roots e^(i2πk/6) for k=0,...,5. So, z=1 (k=0), z=e^(iπ/3) (k=1), z=e^(i2π/3) (k=2), z=-1 (k=3), z=e^(i4π/3) (k=4), z=e^(i5π/3) (k=5). Therefore, z⁴ + z² + 1 = 0 has roots e^(iπ/3), e^(i2π/3), e^(i4π/3), e^(i5π/3). Therefore, these are exactly the 6th roots of unity except for 1 and -1. Therefore, all roots are 6th roots of unity. Therefore, the minimal n is 6. Therefore, the answer is 6?

But let me check again. Suppose n=6. Then, all roots are 6th roots of unity. But is there a smaller n? For example, n=12? No, because 6 is smaller. Wait, but maybe n= something else?

Wait, but let me think again. Suppose that n is the least common multiple of the orders of the roots. The roots are e^(iπ/3), e^(i2π/3), e^(i4π/3), e^(i5π/3). The order of e^(iπ/3) is 6, since (e^(iπ/3))⁶ = e^(i2π) = 1. Similarly, the order of e^(i2π/3) is 3, since (e^(i2π/3))³ = e^(i2π) = 1. Similarly for the others. Therefore, the orders are 3 and 6. Therefore, the least common multiple of 3 and 6 is 6. Therefore, n=6 is the minimal n such that all roots are nth roots of unity. Therefore, the answer is 6. But let me check once more.

Alternatively, maybe I can think of the minimal n such that z⁴ + z² + 1 divides zⁿ - 1. Because if all roots of z⁴ + z² + 1 are nth roots of unity, then z⁴ + z² + 1 divides zⁿ - 1. Therefore, we need to find the minimal n such that z⁴ + z² + 1 divides zⁿ - 1. Since we know that z⁴ + z² + 1 = (z⁶ - 1)/(z² - 1), as we saw earlier. Therefore, z⁶ - 1 = (z² - 1)(z⁴ + z² + 1). Therefore, z⁴ + z² + 1 divides z⁶ - 1. Therefore, if we take n=6, then z⁴ + z² + 1 divides z⁶ - 1, so all roots of z⁴ + z² + 1 are 6th roots of unity. Therefore, n=6 is such a number. Now, is there a smaller n?

Suppose n=3. Then z³ - 1 = (z - 1)(z² + z + 1). But z⁴ + z² + 1 does not divide z³ - 1, since z⁴ + z² + 1 is degree 4 and z³ - 1 is degree 3. Similarly, n=4: z⁴ - 1 = (z² - 1)(z² + 1). Again, z⁴ + z² + 1 is not a factor. For n=5: z⁵ - 1. Not sure, but since the roots are not 5th roots of unity, as we saw earlier, so z⁴ + z² + 1 does not divide z⁵ - 1. For n=6, it divides z⁶ - 1. Therefore, n=6 is the minimal. Therefore, the answer is 6.

But let me check if maybe n=12? Wait, no, since 6 is smaller. But maybe there is a smaller n than 6? Let me check n=2: z² - 1. No. n=3: z³ - 1. No. n=4: z⁴ - 1. No. n=5: z⁵ - 1. No. n=6: yes. Therefore, 6 is the minimal. Therefore, the answer is 6. Therefore, the smallest positive integer n is 6.

But let me check once again with another approach. Let me consider the roots. The roots are e^(iπ/3), e^(i2π/3), e^(i4π/3), e^(i5π/3). Let me write them as complex numbers:

e^(iπ/3) = cos(π/3) + i sin(π/3) = 1/2 + i(√3/2)

Similarly, e^(i2π/3) = -1/2 + i(√3/2)

e^(i4π/3) = -1/2 - i(√3/2)

e^(i5π/3) = 1/2 - i(√3/2)

Now, if these are nth roots of unity, then they must satisfy zⁿ = 1. Let me check for n=6:

For e^(iπ/3): (e^(iπ/3))⁶ = e^(i2π) = 1. Similarly for others. For n=6, they all satisfy z⁶=1. For n=3: e^(iπ/3)³ = e^(iπ) = -1 ≠ 1. Therefore, n=3 is invalid. For n=4: e^(iπ/3)^4 = e^(i4π/3) ≠ 1. Therefore, invalid. For n=5: e^(iπ/3)^5 = e^(i5π/3) ≠ 1. Therefore, invalid. Therefore, n=6 is indeed the minimal. Therefore, the answer is 6.

But wait, let me check if there is a smaller n. For example, n=12? No, since 6 is smaller. Wait, but maybe n= something else? For example, n= 12? No. Wait, but maybe n= something else. Let me think. Suppose that n is the least common multiple of the orders of the roots. The roots have orders 3 and 6. The LCM of 3 and 6 is 6. Therefore, n=6. Therefore, that's the minimal. Therefore, the answer is 6.

Alternatively, maybe I can think of the minimal n such that z⁴ + z² + 1 divides zⁿ - 1. Since z⁴ + z² + 1 divides z⁶ - 1, as we saw earlier, then n=6 is the minimal. Because z⁶ - 1 is the minimal polynomial that has all these roots. Therefore, n=6 is the minimal. Therefore, the answer is 6.

But let me check if there is a smaller n. For example, suppose n= 12? No, since 6 is smaller. Wait, but maybe n= something else. Let me think. Suppose n= 6. Then z⁶ - 1 = (z² - 1)(z⁴ + z² + 1). Therefore, z⁴ + z² + 1 divides z⁶ - 1. Therefore, n=6 is the minimal n. Therefore, the answer is 6.

Therefore, after all these checks, I think the answer is 6. Therefore, the smallest positive integer n is 6.

**Final Answer**
The smallest positive integer $ n $ is \boxed{6}.

To determine the smallest positive integer $ n $ such that all the roots of the polynomial
$$
z^4 + z^2 + 1 = 0
$$
are $ n $th roots of unity, we begin by analyzing the structure of the roots.

---

### Step 1: Factor the Polynomial

Let us make a substitution to simplify the equation. Let $ w = z^2 $. Then the equation becomes:
$$
w^2 + w + 1 = 0
$$
This is a quadratic equation with complex roots:
$$
w = \frac{-1 \pm \sqrt{-3}}{2} = \frac{-1 \pm i\sqrt{3}}{2}
$$
These are the **primitive 3rd roots of unity**, since they are the roots of the cyclotomic polynomial $ \Phi_3(w) = w^2 + w + 1 $.

So, the original equation $ z^4 + z^2 + 1 = 0 $ has roots that are the square roots of these two complex numbers.

---

### Step 2: Find the Roots of the Original Equation

Let’s denote the roots of $ w^2 + w + 1 = 0 $ as:
$$
w_1 = e^{2\pi i / 3}, \quad w_2 = e^{4\pi i / 3}
$$
These are the primitive 3rd roots of unity.

Now, the roots of the original equation are the square roots of $ w_1 $ and $ w_2 $, which are:
$$
z = \pm e^{\pi i / 3}, \quad \pm e^{2\pi i / 3}, \quad \pm e^{4\pi i / 3}, \quad \pm e^{5\pi i / 3}
$$
But since the original equation is of degree 4, we only take **two square roots** of each $ w $, resulting in the four roots:
$$
z = e^{\pi i / 3}, \quad e^{2\pi i / 3}, \quad e^{4\pi i / 3}, \quad e^{5\pi i / 3}
$$

---

### Step 3: Identify the Orders of the Roots

Let’s determine the **order** (i.e., the smallest positive integer $ n $ such that $ z^n = 1 $) of each root:

- $ e^{\pi i / 3} = e^{2\pi i / 6} $: This is a **primitive 6th root of unity**, so its order is 6.
- $ e^{2\pi i / 3} = e^{4\pi i / 6} $: This is a **primitive 3rd root of unity**, so its order is 3.
- $ e^{4\pi i / 3} = e^{8\pi i / 6} $: Also a **primitive 3rd root of unity**, order 3.
- $ e^{5\pi i / 3} = e^{10\pi i / 6} $: A **primitive 6th root of unity**, order 6.

So, the roots have orders 3 and 6.

---

### Step 4: Determine the Smallest $ n $ Such That All Roots Are $ n $th Roots of Unity

To ensure that **all** roots are $ n $th roots of unity, $ n $ must be a multiple of the **least common multiple (LCM)** of the orders of the roots.

- The orders are 3 and 6.
- $ \text{LCM}(3, 6) = 6 $

Thus, the **smallest** $ n $ such that all roots of the equation are $ n $th roots of unity is:
$$
\boxed{6}
$$

---

### Final Verification

We can also observe that:
$$
z^6 - 1 = (z^2 - 1)(z^4 + z^2 + 1)
$$
This shows that the polynomial $ z^4 + z^2 + 1 $ divides $ z^6 - 1 $, and hence all its roots are 6th roots of unity. Since no smaller $ n $ (like 3, 4, or 5) works, we conclude that the minimal such $ n $ is indeed:

$$
\boxed{6}
$$

[Assistant]
Explanation:
Okay, so I need to find the smallest positive integer n such that all the roots of the equation z⁴ + z² + 1 = 0 are nth roots of unity. Hmm, let me start by recalling what roots of unity are. A root of unity is a complex number that satisfies zⁿ = 1 for some positive integer n. So, essentially, these are the solutions to the equation zⁿ = 1. The roots of unity are located on the unit circle in the complex plane and are equally spaced around the circle.

Now, the problem is about the roots of the polynomial z⁴ + z² + 1 = 0. I need to find the smallest n such that each of these roots is an nth root of unity. That means each root should satisfy zⁿ = 1. So, first, maybe I should find the roots of the given polynomial and then determine what n would make them roots of unity.

But solving z⁴ + z² + 1 = 0 directly might be a bit tricky. Let me think. Maybe I can factor this polynomial or relate it to a known equation. Let me try substituting w = z². Then the equation becomes w² + w + 1 = 0. That seems simpler. Let me solve for w first.

So, solving w² + w + 1 = 0. Using the quadratic formula: w = [-1 ± √(1 - 4)] / 2 = [-1 ± √(-3)] / 2 = (-1 ± i√3)/2. Therefore, the solutions for w are complex numbers. Then, since w = z², we can find z by taking square roots of these w's.

So, let me write down the roots. Let me denote the roots of w² + w + 1 = 0 as w₁ and w₂. Then, each root z of the original equation is a square root of w₁ or w₂. Let me compute those square roots.

First, let me recall that the roots of w² + w + 1 = 0 are the primitive third roots of unity. Wait, because the equation w² + w + 1 = 0 is the cyclotomic polynomial Φ₃(w), which factors as (w - e^(2πi/3))(w - e^(4πi/3)). So, these are the primitive 3rd roots of unity. Therefore, w₁ = e^(2πi/3) and w₂ = e^(4πi/3). Therefore, the original equation z⁴ + z² + 1 = 0 has roots which are square roots of these w's.

So, to find z, we need to take square roots of e^(2πi/3) and e^(4πi/3). Let me recall that taking square roots of a complex number on the unit circle involves taking the square root of the modulus (which is 1 here) and halving the angle. So, for example, the square roots of e^(2πi/3) would be e^(πi/3) and e^(πi/3 + πi) = e^(4πi/3). Wait, but hold on, let me check that.

Wait, if I have a complex number e^(iθ), then its square roots are e^(iθ/2) and e^(iθ/2 + πi). So, for θ = 2π/3, the square roots would be e^(iπ/3) and e^(iπ/3 + iπ) = e^(i4π/3). Similarly, for θ = 4π/3, the square roots would be e^(i2π/3) and e^(i2π/3 + iπ) = e^(i5π/3). Therefore, the roots of the original equation z⁴ + z² + 1 = 0 are e^(iπ/3), e^(i4π/3), e^(i2π/3), and e^(i5π/3). Wait, but let me check if that's correct.

Wait, let me verify. Let me compute z⁴ + z² + 1 for z = e^(iπ/3). Let me compute z² first. z² = e^(i2π/3). Then z⁴ = (z²)² = e^(i4π/3). Therefore, z⁴ + z² + 1 = e^(i4π/3) + e^(i2π/3) + 1. Let me compute this sum. Let me recall that e^(i2π/3) + e^(i4π/3) + 1 = 0, because they are the roots of w² + w + 1 = 0. Wait, but here we have e^(i4π/3) + e^(i2π/3) + 1 = 0. Therefore, z⁴ + z² + 1 = 0. So yes, that works. Similarly, for z = e^(i2π/3), z² = e^(i4π/3), z⁴ = e^(i8π/3) = e^(i(8π/3 - 2π)) = e^(i2π/3). Then z⁴ + z² + 1 = e^(i2π/3) + e^(i4π/3) + 1 = 0. So that works. Similarly for the other roots. Therefore, the roots of the original equation are e^(iπ/3), e^(i2π/3), e^(i4π/3), and e^(i5π/3). Wait, but e^(i5π/3) is the same as e^(-iπ/3), right? So, these are the 6th roots of unity? Wait, let me check.

Wait, e^(iπ/3) is a primitive 6th root of unity? Let me recall that the 6th roots of unity are e^(i2πk/6) for k = 0, 1, 2, 3, 4, 5. So, e^(iπ/3) is e^(i2π/6), which is a primitive 6th root of unity. Similarly, e^(i2π/3) is e^(i4π/6), which is a primitive 3rd root of unity. Wait, but 2π/3 is 4π/6. So, e^(i2π/3) is a primitive 3rd root of unity. Similarly, e^(i4π/3) is e^(i8π/6) which is e^(i4π/3), which is also a primitive 3rd root of unity. And e^(i5π/3) is e^(i10π/6) = e^(i5π/3), which is a primitive 6th root of unity. Therefore, the roots of the original equation are two primitive 3rd roots of unity and two primitive 6th roots of unity. Therefore, they are all roots of unity, but with different orders.

Therefore, to find the smallest n such that all these roots are nth roots of unity, we need to find the least common multiple (LCM) of the orders of these roots. The orders of the roots are 3 and 6. The LCM of 3 and 6 is 6. Therefore, n should be 6? Wait, but let me check if all roots are 6th roots of unity. Let me check each root:

- e^(iπ/3): This is e^(i2π/6), so it's a 6th root of unity. Similarly, e^(i5π/3) is also a 6th root of unity. e^(i2π/3) is e^(i4π/6), which is a 3rd root of unity, but since 3 divides 6, it is also a 6th root of unity. Similarly, e^(i4π/3) is e^(i8π/6) = e^(i4π/3), which is also a 6th root of unity. Therefore, all roots are 6th roots of unity. Therefore, n=6?

But wait, let me check if there is a smaller n. For example, n=3? Let me check if all roots are 3rd roots of unity. The roots e^(iπ/3) and e^(i5π/3) are not 3rd roots of unity because their angles are π/3 and 5π/3, which are not multiples of 2π/3. For example, e^(iπ/3) raised to the 3rd power is e^(iπ) = -1 ≠ 1. Therefore, they are not 3rd roots of unity. Therefore, n=3 is too small. Similarly, n=4? Let me check. The roots are e^(iπ/3), e^(i2π/3), e^(i4π/3), e^(i5π/3). Let me check if they are 4th roots of unity. The 4th roots of unity are e^(i0), e^(iπ/2), e^(iπ), e^(i3π/2). None of the roots we have are among these. For example, e^(iπ/3) is not a 4th root of unity. Therefore, n=4 is too small. Similarly, n=5? The 5th roots of unity are e^(i2πk/5) for k=0,...,4. Our roots have angles π/3, 2π/3, 4π/3, 5π/3. These angles are not multiples of 2π/5, so they are not 5th roots of unity. Therefore, n=5 is too small. Therefore, n=6 is the minimal?

Wait, but let me check if there is a smaller n than 6. For example, n=12? Wait, no, we need the minimal n. Wait, but maybe n=6 is the minimal. But let me think again. Let me check if all roots are 6th roots of unity. Since they are, and since there is no smaller n that works, then n=6 is the answer. But let me check if maybe there is a smaller n. For example, n=12? No, that's bigger. Wait, but maybe n= something else?

Alternatively, perhaps I can think algebraically. Let me note that the original equation z⁴ + z² + 1 = 0. Let me try to relate this to cyclotomic polynomials. Let me recall that z⁶ - 1 = (z³ - 1)(z³ + 1) = (z - 1)(z² + z + 1)(z + 1)(z² - z + 1). Also, z⁶ - 1 = (z² - 1)(z⁴ + z² + 1). Wait, let me check that. Let me compute (z² - 1)(z⁴ + z² + 1) = z⁶ + z⁴ + z² - z⁴ - z² - 1 = z⁶ - 1. Yes! Therefore, z⁴ + z² + 1 = (z⁶ - 1)/(z² - 1). Therefore, the roots of z⁴ + z² + 1 = 0 are the roots of z⁶ - 1 = 0 except for the roots of z² - 1 = 0. Therefore, the roots of z⁴ + z² + 1 = 0 are the 6th roots of unity except for z = 1 and z = -1. Therefore, the roots are e^(i2πk/6) for k = 1, 2, 3, 4, 5, excluding k=3 (which gives z=-1) and k=0 (which gives z=1). Wait, let me check:

Wait, z⁶ - 1 = 0 has roots e^(i2πk/6) for k=0,...,5. So, z=1 (k=0), z=e^(iπ/3) (k=1), z=e^(i2π/3) (k=2), z=-1 (k=3), z=e^(i4π/3) (k=4), z=e^(i5π/3) (k=5). Therefore, z⁴ + z² + 1 = 0 has roots e^(iπ/3), e^(i2π/3), e^(i4π/3), e^(i5π/3). Therefore, these are exactly the 6th roots of unity except for 1 and -1. Therefore, all roots are 6th roots of unity. Therefore, the minimal n is 6. Therefore, the answer is 6?

But let me check again. Suppose n=6. Then, all roots are 6th roots of unity. But is there a smaller n? For example, n=12? No, because 6 is smaller. Wait, but maybe n= something else?

Wait, but let me think again. Suppose that n is the least common multiple of the orders of the roots. The roots are e^(iπ/3), e^(i2π/3), e^(i4π/3), e^(i5π/3). The order of e^(iπ/3) is 6, since (e^(iπ/3))⁶ = e^(i2π) = 1. Similarly, the order of e^(i2π/3) is 3, since (e^(i2π/3))³ = e^(i2π) = 1. Similarly for the others. Therefore, the orders are 3 and 6. Therefore, the least common multiple of 3 and 6 is 6. Therefore, n=6 is the minimal n such that all roots are nth roots of unity. Therefore, the answer is 6. But let me check once more.

Alternatively, maybe I can think of the minimal n such that z⁴ + z² + 1 divides zⁿ - 1. Because if all roots of z⁴ + z² + 1 are nth roots of unity, then z⁴ + z² + 1 divides zⁿ - 1. Therefore, we need to find the minimal n such that z⁴ + z² + 1 divides zⁿ - 1. Since we know that z⁴ + z² + 1 = (z⁶ - 1)/(z² - 1), as we saw earlier. Therefore, z⁶ - 1 = (z² - 1)(z⁴ + z² + 1). Therefore, z⁴ + z² + 1 divides z⁶ - 1. Therefore, if we take n=6, then z⁴ + z² + 1 divides z⁶ - 1, so all roots of z⁴ + z² + 1 are 6th roots of unity. Therefore, n=6 is such a number. Now, is there a smaller n?

Suppose n=3. Then z³ - 1 = (z - 1)(z² + z + 1). But z⁴ + z² + 1 does not divide z³ - 1, since z⁴ + z² + 1 is degree 4 and z³ - 1 is degree 3. Similarly, n=4: z⁴ - 1 = (z² - 1)(z² + 1). Again, z⁴ + z² + 1 is not a factor. For n=5: z⁵ - 1. Not sure, but since the roots are not 5th roots of unity, as we saw earlier, so z⁴ + z² + 1 does not divide z⁵ - 1. For n=6, it divides z⁶ - 1. Therefore, n=6 is the minimal. Therefore, the answer is 6.

But let me check if maybe n=12? Wait, no, since 6 is smaller. But maybe there is a smaller n than 6? Let me check n=2: z² - 1. No. n=3: z³ - 1. No. n=4: z⁴ - 1. No. n=5: z⁵ - 1. No. n=6: yes. Therefore, 6 is the minimal. Therefore, the answer is 6. Therefore, the smallest positive integer n is 6.

But let me check once again with another approach. Let me consider the roots. The roots are e^(iπ/3), e^(i2π/3), e^(i4π/3), e^(i5π/3). Let me write them as complex numbers:

e^(iπ/3) = cos(π/3) + i sin(π/3) = 1/2 + i(√3/2)

Similarly, e^(i2π/3) = -1/2 + i(√3/2)

e^(i4π/3) = -1/2 - i(√3/2)

e^(i5π/3) = 1/2 - i(√3/2)

Now, if these are nth roots of unity, then they must satisfy zⁿ = 1. Let me check for n=6:

For e^(iπ/3): (e^(iπ/3))⁶ = e^(i2π) = 1. Similarly for others. For n=6, they all satisfy z⁶=1. For n=3: e^(iπ/3)³ = e^(iπ) = -1 ≠ 1. Therefore, n=3 is invalid. For n=4: e^(iπ/3)^4 = e^(i4π/3) ≠ 1. Therefore, invalid. For n=5: e^(iπ/3)^5 = e^(i5π/3) ≠ 1. Therefore, invalid. Therefore, n=6 is indeed the minimal. Therefore, the answer is 6.

But wait, let me check if there is a smaller n. For example, n=12? No, since 6 is smaller. Wait, but maybe n= something else? For example, n= 12? No. Wait, but maybe n= something else. Let me think. Suppose that n is the least common multiple of the orders of the roots. The roots have orders 3 and 6. The LCM of 3 and 6 is 6. Therefore, n=6. Therefore, that's the minimal. Therefore, the answer is 6.

Alternatively, maybe I can think of the minimal n such that z⁴ + z² + 1 divides zⁿ - 1. Since z⁴ + z² + 1 divides z⁶ - 1, as we saw earlier, then n=6 is the minimal. Because z⁶ - 1 is the minimal polynomial that has all these roots. Therefore, n=6 is the minimal. Therefore, the answer is 6.

But let me check if there is a smaller n. For example, suppose n= 12? No, since 6 is smaller. Wait, but maybe n= something else. Let me think. Suppose n= 6. Then z⁶ - 1 = (z² - 1)(z⁴ + z² + 1). Therefore, z⁴ + z² + 1 divides z⁶ - 1. Therefore, n=6 is the minimal n. Therefore, the answer is 6.

Therefore, after all these checks, I think the answer is 6. Therefore, the smallest positive integer n is 6.

**Final Answer**
The smallest positive integer $ n $ is \boxed{6}.

To determine the smallest positive integer $ n $ such that all the roots of the polynomial
$$
z^4 + z^2 + 1 = 0
$$
are $ n $th roots of unity, we begin by analyzing the structure of the roots.

---

### Step 1: Factor the Polynomial

Let us make a substitution to simplify the equation. Let $ w = z^2 $. Then the equation becomes:
$$
w^2 + w + 1 = 0
$$
This is a quadratic equation with complex roots:
$$
w = \frac{-1 \pm \sqrt{-3}}{2} = \frac{-1 \pm i\sqrt{3}}{2}
$$
These are the **primitive 3rd roots of unity**, since they are the roots of the cyclotomic polynomial $ \Phi_3(w) = w^2 + w + 1 $.

So, the original equation $ z^4 + z^2 + 1 = 0 $ has roots that are the square roots of these two complex numbers.

---

### Step 2: Find the Roots of the Original Equation

Let’s denote the roots of $ w^2 + w + 1 = 0 $ as:
$$
w_1 = e^{2\pi i / 3}, \quad w_2 = e^{4\pi i / 3}
$$
These are the primitive 3rd roots of unity.

Now, the roots of the original equation are the square roots of $ w_1 $ and $ w_2 $, which are:
$$
z = \pm e^{\pi i / 3}, \quad \pm e^{2\pi i / 3}, \quad \pm e^{4\pi i / 3}, \quad \pm e^{5\pi i / 3}
$$
But since the original equation is of degree 4, we only take **two square roots** of each $ w $, resulting in the four roots:
$$
z = e^{\pi i / 3}, \quad e^{2\pi i / 3}, \quad e^{4\pi i / 3}, \quad e^{5\pi i / 3}
$$

---

### Step 3: Identify the Orders of the Roots

Let’s determine the **order** (i.e., the smallest positive integer $ n $ such that $ z^n = 1 $) of each root:

- $ e^{\pi i / 3} = e^{2\pi i / 6} $: This is a **primitive 6th root of unity**, so its order is 6.
- $ e^{2\pi i / 3} = e^{4\pi i / 6} $: This is a **primitive 3rd root of unity**, so its order is 3.
- $ e^{4\pi i / 3} = e^{8\pi i / 6} $: Also a **primitive 3rd root of unity**, order 3.
- $ e^{5\pi i / 3} = e^{10\pi i / 6} $: A **primitive 6th root of unity**, order 6.

So, the roots have orders 3 and 6.

---

### Step 4: Determine the Smallest $ n $ Such That All Roots Are $ n $th Roots of Unity

To ensure that **all** roots are $ n $th roots of unity, $ n $ must be a multiple of the **least common multiple (LCM)** of the orders of the roots.

- The orders are 3 and 6.
- $ \text{LCM}(3, 6) = 6 $

Thus, the **smallest** $ n $ such that all roots of the equation are $ n $th roots of unity is:
$$
\boxed{6}
$$

---

### Final Verification

We can also observe that:
$$
z^6 - 1 = (z^2 - 1)(z^4 + z^2 + 1)
$$
This shows that the polynomial $ z^4 + z^2 + 1 $ divides $ z^6 - 1 $, and hence all its roots are 6th roots of unity. Since no smaller $ n $ (like 3, 4, or 5) works, we conclude that the minimal such $ n $ is indeed:

$$
\boxed{6}
$$

Solution:
Multiplying the equation $z^4 + z^2 + 1 = 0$ by $z^2 - 1 = (z - 1)(z + 1)$, we get $z^6 - 1 = 0$.  Therefore, every root of $z^4 + z^2 + 1 = 0$ is a sixth root of unity.

The sixth roots of unity are $e^{0}$, $e^{2 \pi i/6}$, $e^{4 \pi i/6}$, $e^{6 \pi i/6}$, $e^{8 \pi i/6}$, and $e^{10 \pi i/6}$.  We see that $e^{0} = 1$ and $e^{6 \pi i/6} = e^{\pi i} = -1$, so the roots of
\[z^4 + z^2 + 1 = 0\]are the remaining sixth roots of unity, namely $e^{2 \pi i/6}$, $e^{4 \pi i/6}$, $e^{8 \pi i/6}$, and $e^{10 \pi i/6}$.  The complex number $e^{2 \pi i/6}$ is a primitive sixth root of unity, so by definition, the smallest positive integer $n$ such that $(e^{2 \pi i/6})^n = 1$ is 6.  Therefore, the smallest possible value of $n$ is $\boxed{6}$.


Explanation:
Okay, so I need to figure out what power of 4 equals 8. Hmm, let me think. The question is asking for an exponent such that when I raise 4 to that exponent, I get 8. In other words, find x where 4^x = 8. Alright, let me recall how exponents work.

First, I know that 4 is 2 squared, and 8 is 2 cubed. Maybe I can express both numbers with the same base? Let me try that. If I rewrite 4 as 2^2, then 4^x would be (2^2)^x. Using the exponent rule (a^b)^c = a^(b*c), that becomes 2^(2x). So, 2^(2x) = 8. But 8 is 2^3, right? So substituting that in, I get 2^(2x) = 2^3.

Since the bases are the same, the exponents must be equal. Therefore, 2x = 3. Solving for x, I divide both sides by 2, so x = 3/2. Let me check if that works. If I take 4^(3/2), that should be equal to 8. Let me compute that.

First, 4^(1/2) is the square root of 4, which is 2. Then, 4^(3/2) is the same as (4^(1/2))^3, which is 2^3 = 8. Yep, that checks out. So x is 3/2.

Wait, let me make sure there's no other way to approach this. Maybe using logarithms? Let me try that method too. Starting with 4^x = 8. Taking the logarithm of both sides. Let's use natural logarithm for this. ln(4^x) = ln(8). Then, using the power rule, x*ln(4) = ln(8). Therefore, x = ln(8)/ln(4).

Calculating that, ln(8) is ln(2^3) = 3*ln(2), and ln(4) is ln(2^2) = 2*ln(2). So substituting back, x = (3*ln(2))/(2*ln(2)) = 3/2. The ln(2) cancels out, leaving 3/2. So that's the same answer as before.

Another way to think about it: since 4 and 8 are both powers of 2, converting them to the same base makes it easier. So 4 is 2 squared, 8 is 2 cubed. So if I have 4^x = 8, then (2^2)^x = 2^3, which simplifies to 2^(2x) = 2^3, so 2x = 3, so x = 3/2. Yep, same result.

Is there a way to do this without converting to base 2? Maybe using logarithms with base 4? Let me try. If I take log base 4 of both sides of 4^x = 8, then log base 4 of 4^x is x, and log base 4 of 8 is the answer. So x = log_4(8). How do I compute that? Well, log base 4 of 8 is the exponent you need to raise 4 to get 8. Which is exactly what we're trying to find. But maybe I can express 8 as a power of 4? Let me see. 4^1 = 4, 4^2 = 16. So 8 is between 4^1 and 4^2. So it's not an integer exponent. But maybe express it in terms of exponents?

Alternatively, using change of base formula. log_4(8) = ln(8)/ln(4) or log(8)/log(4), which is the same as before. Which again gives 3/2. So regardless of the method, I get 3/2.

Let me verify once more with decimal exponents. Let me compute 4^(3/2). 3/2 is 1.5. So 4^1.5. Let me compute that. 4^1 is 4, 4^0.5 is 2, so multiplying them together, 4 * 2 = 8. Exactly. So that's correct.

Alternatively, if I use a calculator, 4^(3/2) is sqrt(4)^3 = 2^3 = 8. Yep.

Is there any chance I could have messed up the exponent rules? Let me think. For example, if I thought that 4^x = 8 implies x = log_4(8), which is correct. Then, if I didn't remember how to compute that, but I know that 4 is 2 squared and 8 is 2 cubed, so maybe I can write log_4(8) as log_2(8)/log_2(4) by change of base formula. Which is 3/2. Yep.

Alternatively, if I didn't remember the change of base formula, but I know that log_b(a) = c means that b^c = a. So if I have log_4(8) = x, then 4^x = 8. Which is the original equation. So that's circular. But if I can express 8 as 4^(something), then that something is x. Since 4^1 = 4, 4^2 = 16, so 8 is 4^(1 + something). Let me write 8 as 4 * 2. So 4^1 * 2^1. But 2 is 4^(1/2). So 4^1 * 4^(1/2) = 4^(1 + 1/2) = 4^(3/2). Therefore, 8 = 4^(3/2). Therefore, x is 3/2.

So all methods lead to the same answer. Therefore, I think the answer is 3/2.

Wait, the question says "Express your answer as a common fraction." So 3/2 is already a common fraction. So I think that's the answer.

Just to recap, the problem is asking for the exponent x such that 4^x = 8. By expressing both numbers as powers of 2, we can equate the exponents and solve for x. Alternatively, using logarithms or change of base formula also leads to the same result. All methods confirm that x is 3/2. Therefore, the answer is 3/2.

**Final Answer**
The power of 4 that equals 8 is \boxed{\dfrac{3}{2}}.

To determine what power of 4 equals 8, we are essentially solving the equation:

$$
4^x = 8
$$

---

### Step 1: Express Both Numbers with the Same Base

We know that:

- $ 4 = 2^2 $
- $ 8 = 2^3 $

So we can rewrite the original equation using base 2:

$$
(2^2)^x = 2^3
$$

---

### Step 2: Apply Exponent Rules

Using the rule $ (a^b)^c = a^{bc} $, the left-hand side becomes:

$$
2^{2x} = 2^3
$$

Since the bases are the same, we can equate the exponents:

$$
2x = 3
$$

---

### Step 3: Solve for $ x $

$$
x = \frac{3}{2}
$$

---

### Step 4: Verify the Solution

Let’s verify that $ 4^{3/2} = 8 $:

$$
4^{3/2} = (4^{1/2})^3 = (\sqrt{4})^3 = 2^3 = 8
$$

This confirms that the solution is correct.

---

### Final Answer

$$
\boxed{\dfrac{3}{2}}
$$

[Assistant]
Explanation:
Okay, so I need to figure out what power of 4 equals 8. Hmm, let me think. The question is asking for an exponent such that when I raise 4 to that exponent, I get 8. In other words, find x where 4^x = 8. Alright, let me recall how exponents work.

First, I know that 4 is 2 squared, and 8 is 2 cubed. Maybe I can express both numbers with the same base? Let me try that. If I rewrite 4 as 2^2, then 4^x would be (2^2)^x. Using the exponent rule (a^b)^c = a^(b*c), that becomes 2^(2x). So, 2^(2x) = 8. But 8 is 2^3, right? So substituting that in, I get 2^(2x) = 2^3.

Since the bases are the same, the exponents must be equal. Therefore, 2x = 3. Solving for x, I divide both sides by 2, so x = 3/2. Let me check if that works. If I take 4^(3/2), that should be equal to 8. Let me compute that.

First, 4^(1/2) is the square root of 4, which is 2. Then, 4^(3/2) is the same as (4^(1/2))^3, which is 2^3 = 8. Yep, that checks out. So x is 3/2.

Wait, let me make sure there's no other way to approach this. Maybe using logarithms? Let me try that method too. Starting with 4^x = 8. Taking the logarithm of both sides. Let's use natural logarithm for this. ln(4^x) = ln(8). Then, using the power rule, x*ln(4) = ln(8). Therefore, x = ln(8)/ln(4).

Calculating that, ln(8) is ln(2^3) = 3*ln(2), and ln(4) is ln(2^2) = 2*ln(2). So substituting back, x = (3*ln(2))/(2*ln(2)) = 3/2. The ln(2) cancels out, leaving 3/2. So that's the same answer as before.

Another way to think about it: since 4 and 8 are both powers of 2, converting them to the same base makes it easier. So 4 is 2 squared, 8 is 2 cubed. So if I have 4^x = 8, then (2^2)^x = 2^3, which simplifies to 2^(2x) = 2^3, so 2x = 3, so x = 3/2. Yep, same result.

Is there a way to do this without converting to base 2? Maybe using logarithms with base 4? Let me try. If I take log base 4 of both sides of 4^x = 8, then log base 4 of 4^x is x, and log base 4 of 8 is the answer. So x = log_4(8). How do I compute that? Well, log base 4 of 8 is the exponent you need to raise 4 to get 8. Which is exactly what we're trying to find. But maybe I can express 8 as a power of 4? Let me see. 4^1 = 4, 4^2 = 16. So 8 is between 4^1 and 4^2. So it's not an integer exponent. But maybe express it in terms of exponents?

Alternatively, using change of base formula. log_4(8) = ln(8)/ln(4) or log(8)/log(4), which is the same as before. Which again gives 3/2. So regardless of the method, I get 3/2.

Let me verify once more with decimal exponents. Let me compute 4^(3/2). 3/2 is 1.5. So 4^1.5. Let me compute that. 4^1 is 4, 4^0.5 is 2, so multiplying them together, 4 * 2 = 8. Exactly. So that's correct.

Alternatively, if I use a calculator, 4^(3/2) is sqrt(4)^3 = 2^3 = 8. Yep.

Is there any chance I could have messed up the exponent rules? Let me think. For example, if I thought that 4^x = 8 implies x = log_4(8), which is correct. Then, if I didn't remember how to compute that, but I know that 4 is 2 squared and 8 is 2 cubed, so maybe I can write log_4(8) as log_2(8)/log_2(4) by change of base formula. Which is 3/2. Yep.

Alternatively, if I didn't remember the change of base formula, but I know that log_b(a) = c means that b^c = a. So if I have log_4(8) = x, then 4^x = 8. Which is the original equation. So that's circular. But if I can express 8 as 4^(something), then that something is x. Since 4^1 = 4, 4^2 = 16, so 8 is 4^(1 + something). Let me write 8 as 4 * 2. So 4^1 * 2^1. But 2 is 4^(1/2). So 4^1 * 4^(1/2) = 4^(1 + 1/2) = 4^(3/2). Therefore, 8 = 4^(3/2). Therefore, x is 3/2.

So all methods lead to the same answer. Therefore, I think the answer is 3/2.

Wait, the question says "Express your answer as a common fraction." So 3/2 is already a common fraction. So I think that's the answer.

Just to recap, the problem is asking for the exponent x such that 4^x = 8. By expressing both numbers as powers of 2, we can equate the exponents and solve for x. Alternatively, using logarithms or change of base formula also leads to the same result. All methods confirm that x is 3/2. Therefore, the answer is 3/2.

**Final Answer**
The power of 4 that equals 8 is \boxed{\dfrac{3}{2}}.

To determine what power of 4 equals 8, we are essentially solving the equation:

$$
4^x = 8
$$

---

### Step 1: Express Both Numbers with the Same Base

We know that:

- $ 4 = 2^2 $
- $ 8 = 2^3 $

So we can rewrite the original equation using base 2:

$$
(2^2)^x = 2^3
$$

---

### Step 2: Apply Exponent Rules

Using the rule $ (a^b)^c = a^{bc} $, the left-hand side becomes:

$$
2^{2x} = 2^3
$$

Since the bases are the same, we can equate the exponents:

$$
2x = 3
$$

---

### Step 3: Solve for $ x $

$$
x = \frac{3}{2}
$$

---

### Step 4: Verify the Solution

Let’s verify that $ 4^{3/2} = 8 $:

$$
4^{3/2} = (4^{1/2})^3 = (\sqrt{4})^3 = 2^3 = 8
$$

This confirms that the solution is correct.

---

### Final Answer

$$
\boxed{\dfrac{3}{2}}
$$

Solution: We are asked to solve $4^x=8$ for $x$.  Writing $4$ as $2^2$ and $8$ as $2^3$, the equation becomes $(2^2)^x=2^3$.  The left-hand side simplifies to $2^{2x}$, so we may set exponents equal to find $2x=3$, which implies $x=\boxed{\frac{3}{2}}$.


Explanation:
Okay, so I need to find the quotient when x⁶ - 3 is divided by x + 1. Hmm, let me think. I remember that dividing polynomials can be done using long division or maybe synthetic division. Since the divisor is a linear polynomial, x + 1, synthetic division might be easier. But wait, let me recall how synthetic division works.

First, synthetic division is used when dividing by a linear factor of the form x - c. In this case, the divisor is x + 1, which can be written as x - (-1). So, c would be -1. That seems right. So, I can use synthetic division with c = -1.

But before I jump into synthetic division, maybe I should check if there's another method. Alternatively, I could use polynomial long division. Let me try both methods and see which one is more straightforward.

Starting with synthetic division. Let me set up the coefficients of the polynomial x⁶ - 3. The polynomial is x⁶ + 0x⁵ + 0x⁴ + 0x³ + 0x² + 0x - 3. So, the coefficients are 1, 0, 0, 0, 0, 0, -3.

Now, synthetic division steps:

1. Write down the coefficients: 1, 0, 0, 0, 0, 0, -3.
2. Use c = -1.
3. Bring down the leading coefficient.
4. Multiply by c and add to the next coefficient.
5. Repeat until all coefficients are processed.

Let me write this out step by step.

First, set up the coefficients:

-1 | 1   0   0   0   0   0   -3

Start with bringing down the 1.

1

Multiply 1 by -1: -1. Add to the next coefficient (0): 0 + (-1) = -1.

Now, we have:

1   -1

Multiply -1 by -1: 1. Add to next coefficient (0): 0 + 1 = 1.

Now:

1   -1   1

Multiply 1 by -1: -1. Add to next coefficient (0): 0 + (-1) = -1.

Now:

1   -1   1   -1

Multiply -1 by -1: 1. Add to next coefficient (0): 0 + 1 = 1.

Now:

1   -1   1   -1   1

Multiply 1 by -1: -1. Add to next coefficient (0): 0 + (-1) = -1.

Now:

1   -1   1   -1   1   -1

Multiply -1 by -1: 1. Add to next coefficient (-3): -3 + 1 = -2.

So the final row is:

1   -1   1   -1   1   -1   -2

Therefore, the quotient is the polynomial formed by these coefficients, except the last one which is the remainder. So the quotient is x⁵ - x⁴ + x³ - x² + x - 1, and the remainder is -2.

Wait, but the question asks for the quotient, so the answer is x⁵ - x⁴ + x³ - x² + x - 1. Let me check if this is correct.

Alternatively, maybe I can verify by multiplying the divisor (x + 1) by the quotient and adding the remainder to see if I get the original polynomial.

Let me compute (x + 1)(x⁵ - x⁴ + x³ - x² + x - 1) + (-2).

First, multiply x by each term of the quotient:

x * x⁵ = x⁶

x * (-x⁴) = -x⁵

x * x³ = x⁴

x * (-x²) = -x³

x * x = x²

x * (-1) = -x

Then multiply 1 by each term of the quotient:

1 * x⁵ = x⁵

1 * (-x⁴) = -x⁴

1 * x³ = x³

1 * (-x²) = -x²

1 * x = x

1 * (-1) = -1

Now, add all these terms together:

x⁶ - x⁵ + x⁴ - x³ + x² - x + x⁵ - x⁴ + x³ - x² + x - 1

Now combine like terms:

x⁶ + (-x⁵ + x⁵) + (x⁴ - x⁴) + (-x³ + x³) + (x² - x²) + (-x + x) - 1

Simplify each term:

x⁶ + 0 + 0 + 0 + 0 + 0 - 1 = x⁶ - 1

But the original polynomial is x⁶ - 3. So, according to this, (x + 1)(quotient) + remainder = x⁶ - 1 + (-2) = x⁶ - 3. Wait, that's correct! Because the remainder is -2. So, the calculation checks out. Therefore, the quotient is indeed x⁵ - x⁴ + x³ - x² + x - 1.

Alternatively, maybe I can use polynomial long division to confirm. Let me try that method as well.

Dividing x⁶ - 3 by x + 1.

First, divide x⁶ by x, which gives x⁵. Multiply (x + 1) by x⁵: x⁶ + x⁵. Subtract this from the original polynomial:

(x⁶ - 3) - (x⁶ + x⁵) = -x⁵ - 3.

Next, bring down the next term, but since there are no other terms, we just have -x⁵ - 3. Now divide -x⁵ by x, which is -x⁴. Multiply (x + 1) by -x⁴: -x⁵ - x⁴. Subtract this from the current polynomial:

(-x⁵ - 3) - (-x⁵ - x⁴) = 0 + x⁴ - 3 = x⁴ - 3.

Next, divide x⁴ by x to get x³. Multiply (x + 1) by x³: x⁴ + x³. Subtract:

(x⁴ - 3) - (x⁴ + x³) = -x³ - 3.

Divide -x³ by x to get -x². Multiply (x + 1) by -x²: -x³ - x². Subtract:

(-x³ - 3) - (-x³ - x²) = 0 + x² - 3 = x² - 3.

Divide x² by x to get x. Multiply (x + 1) by x: x² + x. Subtract:

(x² - 3) - (x² + x) = -x - 3.

Divide -x by x to get -1. Multiply (x + 1) by -1: -x -1. Subtract:

(-x - 3) - (-x -1) = 0 - 2 = -2.

So the remainder is -2, and the quotient is x⁵ - x⁴ + x³ - x² + x -1. Which matches the result from synthetic division.

Therefore, the quotient is x⁵ - x⁴ + x³ - x² + x -1.

Alternatively, maybe there's another way to think about this? For example, using the Remainder Theorem? Wait, the Remainder Theorem says that the remainder of dividing by x - c is f(c). But here, we are dividing by x + 1, so c = -1. So, the remainder is f(-1). Let me check that.

f(x) = x⁶ - 3. Then f(-1) = (-1)^6 - 3 = 1 - 3 = -2. Which matches the remainder we found earlier. So that's consistent.

But since the question asks for the quotient, not the remainder, so that's just extra confirmation.

Alternatively, maybe I can factor x⁶ - 3? But x⁶ - 3 is not a standard difference of squares or cubes. Wait, x⁶ is (x³)², so maybe x⁶ - 3 = (x³)^2 - (√3)^2, which would factor as (x³ - √3)(x³ + √3). But that might not help with dividing by x + 1.

Alternatively, maybe using polynomial division with substitution? For example, if I let y = x + 1, but that might complicate things.

Alternatively, maybe using the fact that x⁶ - 1 factors as (x + 1)(x⁵ - x⁴ + x³ - x² + x - 1) + 0, but since our polynomial is x⁶ - 3, which is x⁶ - 1 - 2, so that would be (x + 1)(quotient) - 2. Which is consistent with our previous result.

So, all methods point to the quotient being x⁵ - x⁴ + x³ - x² + x - 1.

Wait, just to make sure, let me check with another value. Let me pick a value for x, compute f(x) and see if (x + 1)*quotient + remainder equals f(x).

Let me choose x = 0.

Original polynomial: 0⁶ - 3 = -3.

Divisor: 0 + 1 = 1.

Quotient at x=0: 0⁵ - 0⁴ + 0³ - 0² + 0 -1 = -1.

Remainder is -2.

So, (0 + 1)*(-1) + (-2) = -1 -2 = -3. Which matches f(0) = -3.

Another test: x = 1.

Original polynomial: 1⁶ - 3 = 1 - 3 = -2.

Divisor: 1 + 1 = 2.

Quotient at x=1: 1 - 1 + 1 - 1 + 1 -1 = 0.

Remainder is -2.

So, (1 + 1)*0 + (-2) = 0 -2 = -2. Which matches f(1) = -2.

Another test: x = 2.

Original polynomial: 64 - 3 = 61.

Divisor: 2 + 1 = 3.

Quotient at x=2: 32 - 16 + 8 - 4 + 2 -1 = 32 -16 is 16, +8 is 24, -4 is 20, +2 is 22, -1 is 21.

Remainder is -2.

So, (2 + 1)*21 + (-2) = 63 - 2 = 61. Which matches f(2) = 61.

So, all tests pass. Therefore, the quotient is indeed x⁵ - x⁴ + x³ - x² + x -1.

Therefore, the answer is x⁵ - x⁴ + x³ - x² + x -1.

But let me check if there's a simpler way to write this. For example, maybe factoring or recognizing a pattern.

Looking at the quotient: x⁵ - x⁴ + x³ - x² + x -1.

This seems like an alternating sum of powers of x. Let me see:

It can be written as x⁵ - x⁴ + x³ - x² + x -1 = x⁵ + x³ + x - (x⁴ + x² + 1).

Alternatively, maybe factor it? Let me try.

But since we already know that x + 1 is a factor of x⁶ - 1, and x⁶ - 3 = (x⁶ - 1) - 2, so the quotient is the same as (x⁶ - 1)/(x + 1) minus 2/(x + 1). But since we are dividing by x + 1, the quotient is (x⁶ - 1)/(x + 1) with remainder -2. But (x⁶ - 1)/(x + 1) is the quotient we found earlier, which is x⁵ - x⁴ + x³ - x² + x -1.

Alternatively, since x⁶ - 1 factors as (x + 1)(x⁵ - x⁴ + x³ - x² + x -1). Therefore, the quotient is indeed that polynomial.

Therefore, I think I can be confident that the quotient is x⁵ - x⁴ + x³ - x² + x -1.

Just to recap, the steps were:

1. Recognize that dividing by x + 1 can be done via synthetic division with c = -1.

2. Set up coefficients of the polynomial x⁶ - 3, which are 1, 0, 0, 0, 0, 0, -3.

3. Perform synthetic division step by step, resulting in coefficients of the quotient and remainder.

4. Verified the result by multiplying the divisor by the quotient and adding the remainder to get back the original polynomial.

5. Checked with specific values of x to ensure correctness.

Therefore, the quotient is x⁵ - x⁴ + x³ - x² + x -1.

**Final Answer**
The quotient is \boxed{x^5 - x^4 + x^3 - x^2 + x - 1}.

To find the quotient when $ x^6 - 3 $ is divided by $ x + 1 $, we can use **synthetic division**, which is a streamlined method for dividing a polynomial by a linear factor of the form $ x - c $.

---

### Step 1: Identify the divisor and set up synthetic division

We are dividing by $ x + 1 $, which can be rewritten as $ x - (-1) $. So, the value of $ c $ is $ -1 $.

The polynomial to divide is:

$$
x^6 - 3 = x^6 + 0x^5 + 0x^4 + 0x^3 + 0x^2 + 0x - 3
$$

So the coefficients are:

$$
1, 0, 0, 0, 0, 0, -3
$$

---

### Step 2: Perform synthetic division

We set up the synthetic division as follows:

$$
\begin{array}{r|rrrrrrr}
-1 & 1 & 0 & 0 & 0 & 0 & 0 & -3 \\
   &   & -1 & 1 & -1 & 1 & -1 & 1 \\
\hline
   & 1 & -1 & 1 & -1 & 1 & -1 & -2 \\
\end{array}
$$

- Bring down the first coefficient: $ 1 $
- Multiply $ 1 \times -1 = -1 $, add to next coefficient: $ 0 + (-1) = -1 $
- Multiply $ -1 \times -1 = 1 $, add to next coefficient: $ 0 + 1 = 1 $
- Multiply $ 1 \times -1 = -1 $, add to next coefficient: $ 0 + (-1) = -1 $
- Multiply $ -1 \times -1 = 1 $, add to next coefficient: $ 0 + 1 = 1 $
- Multiply $ 1 \times -1 = -1 $, add to next coefficient: $ 0 + (-1) = -1 $
- Multiply $ -1 \times -1 = 1 $, add to last coefficient: $ -3 + 1 = -2 $

---

### Step 3: Interpret the result

The final row of the synthetic division gives the coefficients of the **quotient** and the **remainder**:

- Quotient: $ x^5 - x^4 + x^3 - x^2 + x - 1 $
- Remainder: $ -2 $

---

### Step 4: Verification

To confirm, we can multiply the divisor $ x + 1 $ by the quotient and add the remainder:

$$
(x + 1)(x^5 - x^4 + x^3 - x^2 + x - 1) + (-2)
$$

Expanding the product:

$$
x^6 - x^5 + x^4 - x^3 + x^2 - x + x^5 - x^4 + x^3 - x^2 + x - 1
$$

Combine like terms:

$$
x^6 + (-x^5 + x^5) + (x^4 - x^4) + (-x^3 + x^3) + (x^2 - x^2) + (-x + x) - 1 = x^6 - 1
$$

Add the remainder $ -2 $:

$$
x^6 - 1 - 2 = x^6 - 3
$$

This matches the original polynomial, confirming the result.

---

### Final Answer

$$
\boxed{x^5 - x^4 + x^3 - x^2 + x - 1}
$$

[Assistant]
Explanation:
Okay, so I need to find the quotient when x⁶ - 3 is divided by x + 1. Hmm, let me think. I remember that dividing polynomials can be done using long division or maybe synthetic division. Since the divisor is a linear polynomial, x + 1, synthetic division might be easier. But wait, let me recall how synthetic division works.

First, synthetic division is used when dividing by a linear factor of the form x - c. In this case, the divisor is x + 1, which can be written as x - (-1). So, c would be -1. That seems right. So, I can use synthetic division with c = -1.

But before I jump into synthetic division, maybe I should check if there's another method. Alternatively, I could use polynomial long division. Let me try both methods and see which one is more straightforward.

Starting with synthetic division. Let me set up the coefficients of the polynomial x⁶ - 3. The polynomial is x⁶ + 0x⁵ + 0x⁴ + 0x³ + 0x² + 0x - 3. So, the coefficients are 1, 0, 0, 0, 0, 0, -3.

Now, synthetic division steps:

1. Write down the coefficients: 1, 0, 0, 0, 0, 0, -3.
2. Use c = -1.
3. Bring down the leading coefficient.
4. Multiply by c and add to the next coefficient.
5. Repeat until all coefficients are processed.

Let me write this out step by step.

First, set up the coefficients:

-1 | 1   0   0   0   0   0   -3

Start with bringing down the 1.

1

Multiply 1 by -1: -1. Add to the next coefficient (0): 0 + (-1) = -1.

Now, we have:

1   -1

Multiply -1 by -1: 1. Add to next coefficient (0): 0 + 1 = 1.

Now:

1   -1   1

Multiply 1 by -1: -1. Add to next coefficient (0): 0 + (-1) = -1.

Now:

1   -1   1   -1

Multiply -1 by -1: 1. Add to next coefficient (0): 0 + 1 = 1.

Now:

1   -1   1   -1   1

Multiply 1 by -1: -1. Add to next coefficient (0): 0 + (-1) = -1.

Now:

1   -1   1   -1   1   -1

Multiply -1 by -1: 1. Add to next coefficient (-3): -3 + 1 = -2.

So the final row is:

1   -1   1   -1   1   -1   -2

Therefore, the quotient is the polynomial formed by these coefficients, except the last one which is the remainder. So the quotient is x⁵ - x⁴ + x³ - x² + x - 1, and the remainder is -2.

Wait, but the question asks for the quotient, so the answer is x⁵ - x⁴ + x³ - x² + x - 1. Let me check if this is correct.

Alternatively, maybe I can verify by multiplying the divisor (x + 1) by the quotient and adding the remainder to see if I get the original polynomial.

Let me compute (x + 1)(x⁵ - x⁴ + x³ - x² + x - 1) + (-2).

First, multiply x by each term of the quotient:

x * x⁵ = x⁶

x * (-x⁴) = -x⁵

x * x³ = x⁴

x * (-x²) = -x³

x * x = x²

x * (-1) = -x

Then multiply 1 by each term of the quotient:

1 * x⁵ = x⁵

1 * (-x⁴) = -x⁴

1 * x³ = x³

1 * (-x²) = -x²

1 * x = x

1 * (-1) = -1

Now, add all these terms together:

x⁶ - x⁵ + x⁴ - x³ + x² - x + x⁵ - x⁴ + x³ - x² + x - 1

Now combine like terms:

x⁶ + (-x⁵ + x⁵) + (x⁴ - x⁴) + (-x³ + x³) + (x² - x²) + (-x + x) - 1

Simplify each term:

x⁶ + 0 + 0 + 0 + 0 + 0 - 1 = x⁶ - 1

But the original polynomial is x⁶ - 3. So, according to this, (x + 1)(quotient) + remainder = x⁶ - 1 + (-2) = x⁶ - 3. Wait, that's correct! Because the remainder is -2. So, the calculation checks out. Therefore, the quotient is indeed x⁵ - x⁴ + x³ - x² + x - 1.

Alternatively, maybe I can use polynomial long division to confirm. Let me try that method as well.

Dividing x⁶ - 3 by x + 1.

First, divide x⁶ by x, which gives x⁵. Multiply (x + 1) by x⁵: x⁶ + x⁵. Subtract this from the original polynomial:

(x⁶ - 3) - (x⁶ + x⁵) = -x⁵ - 3.

Next, bring down the next term, but since there are no other terms, we just have -x⁵ - 3. Now divide -x⁵ by x, which is -x⁴. Multiply (x + 1) by -x⁴: -x⁵ - x⁴. Subtract this from the current polynomial:

(-x⁵ - 3) - (-x⁵ - x⁴) = 0 + x⁴ - 3 = x⁴ - 3.

Next, divide x⁴ by x to get x³. Multiply (x + 1) by x³: x⁴ + x³. Subtract:

(x⁴ - 3) - (x⁴ + x³) = -x³ - 3.

Divide -x³ by x to get -x². Multiply (x + 1) by -x²: -x³ - x². Subtract:

(-x³ - 3) - (-x³ - x²) = 0 + x² - 3 = x² - 3.

Divide x² by x to get x. Multiply (x + 1) by x: x² + x. Subtract:

(x² - 3) - (x² + x) = -x - 3.

Divide -x by x to get -1. Multiply (x + 1) by -1: -x -1. Subtract:

(-x - 3) - (-x -1) = 0 - 2 = -2.

So the remainder is -2, and the quotient is x⁵ - x⁴ + x³ - x² + x -1. Which matches the result from synthetic division.

Therefore, the quotient is x⁵ - x⁴ + x³ - x² + x -1.

Alternatively, maybe there's another way to think about this? For example, using the Remainder Theorem? Wait, the Remainder Theorem says that the remainder of dividing by x - c is f(c). But here, we are dividing by x + 1, so c = -1. So, the remainder is f(-1). Let me check that.

f(x) = x⁶ - 3. Then f(-1) = (-1)^6 - 3 = 1 - 3 = -2. Which matches the remainder we found earlier. So that's consistent.

But since the question asks for the quotient, not the remainder, so that's just extra confirmation.

Alternatively, maybe I can factor x⁶ - 3? But x⁶ - 3 is not a standard difference of squares or cubes. Wait, x⁶ is (x³)², so maybe x⁶ - 3 = (x³)^2 - (√3)^2, which would factor as (x³ - √3)(x³ + √3). But that might not help with dividing by x + 1.

Alternatively, maybe using polynomial division with substitution? For example, if I let y = x + 1, but that might complicate things.

Alternatively, maybe using the fact that x⁶ - 1 factors as (x + 1)(x⁵ - x⁴ + x³ - x² + x - 1) + 0, but since our polynomial is x⁶ - 3, which is x⁶ - 1 - 2, so that would be (x + 1)(quotient) - 2. Which is consistent with our previous result.

So, all methods point to the quotient being x⁵ - x⁴ + x³ - x² + x - 1.

Wait, just to make sure, let me check with another value. Let me pick a value for x, compute f(x) and see if (x + 1)*quotient + remainder equals f(x).

Let me choose x = 0.

Original polynomial: 0⁶ - 3 = -3.

Divisor: 0 + 1 = 1.

Quotient at x=0: 0⁵ - 0⁴ + 0³ - 0² + 0 -1 = -1.

Remainder is -2.

So, (0 + 1)*(-1) + (-2) = -1 -2 = -3. Which matches f(0) = -3.

Another test: x = 1.

Original polynomial: 1⁶ - 3 = 1 - 3 = -2.

Divisor: 1 + 1 = 2.

Quotient at x=1: 1 - 1 + 1 - 1 + 1 -1 = 0.

Remainder is -2.

So, (1 + 1)*0 + (-2) = 0 -2 = -2. Which matches f(1) = -2.

Another test: x = 2.

Original polynomial: 64 - 3 = 61.

Divisor: 2 + 1 = 3.

Quotient at x=2: 32 - 16 + 8 - 4 + 2 -1 = 32 -16 is 16, +8 is 24, -4 is 20, +2 is 22, -1 is 21.

Remainder is -2.

So, (2 + 1)*21 + (-2) = 63 - 2 = 61. Which matches f(2) = 61.

So, all tests pass. Therefore, the quotient is indeed x⁵ - x⁴ + x³ - x² + x -1.

Therefore, the answer is x⁵ - x⁴ + x³ - x² + x -1.

But let me check if there's a simpler way to write this. For example, maybe factoring or recognizing a pattern.

Looking at the quotient: x⁵ - x⁴ + x³ - x² + x -1.

This seems like an alternating sum of powers of x. Let me see:

It can be written as x⁵ - x⁴ + x³ - x² + x -1 = x⁵ + x³ + x - (x⁴ + x² + 1).

Alternatively, maybe factor it? Let me try.

But since we already know that x + 1 is a factor of x⁶ - 1, and x⁶ - 3 = (x⁶ - 1) - 2, so the quotient is the same as (x⁶ - 1)/(x + 1) minus 2/(x + 1). But since we are dividing by x + 1, the quotient is (x⁶ - 1)/(x + 1) with remainder -2. But (x⁶ - 1)/(x + 1) is the quotient we found earlier, which is x⁵ - x⁴ + x³ - x² + x -1.

Alternatively, since x⁶ - 1 factors as (x + 1)(x⁵ - x⁴ + x³ - x² + x -1). Therefore, the quotient is indeed that polynomial.

Therefore, I think I can be confident that the quotient is x⁵ - x⁴ + x³ - x² + x -1.

Just to recap, the steps were:

1. Recognize that dividing by x + 1 can be done via synthetic division with c = -1.

2. Set up coefficients of the polynomial x⁶ - 3, which are 1, 0, 0, 0, 0, 0, -3.

3. Perform synthetic division step by step, resulting in coefficients of the quotient and remainder.

4. Verified the result by multiplying the divisor by the quotient and adding the remainder to get back the original polynomial.

5. Checked with specific values of x to ensure correctness.

Therefore, the quotient is x⁵ - x⁴ + x³ - x² + x -1.

**Final Answer**
The quotient is \boxed{x^5 - x^4 + x^3 - x^2 + x - 1}.

To find the quotient when $ x^6 - 3 $ is divided by $ x + 1 $, we can use **synthetic division**, which is a streamlined method for dividing a polynomial by a linear factor of the form $ x - c $.

---

### Step 1: Identify the divisor and set up synthetic division

We are dividing by $ x + 1 $, which can be rewritten as $ x - (-1) $. So, the value of $ c $ is $ -1 $.

The polynomial to divide is:

$$
x^6 - 3 = x^6 + 0x^5 + 0x^4 + 0x^3 + 0x^2 + 0x - 3
$$

So the coefficients are:

$$
1, 0, 0, 0, 0, 0, -3
$$

---

### Step 2: Perform synthetic division

We set up the synthetic division as follows:

$$
\begin{array}{r|rrrrrrr}
-1 & 1 & 0 & 0 & 0 & 0 & 0 & -3 \\
   &   & -1 & 1 & -1 & 1 & -1 & 1 \\
\hline
   & 1 & -1 & 1 & -1 & 1 & -1 & -2 \\
\end{array}
$$

- Bring down the first coefficient: $ 1 $
- Multiply $ 1 \times -1 = -1 $, add to next coefficient: $ 0 + (-1) = -1 $
- Multiply $ -1 \times -1 = 1 $, add to next coefficient: $ 0 + 1 = 1 $
- Multiply $ 1 \times -1 = -1 $, add to next coefficient: $ 0 + (-1) = -1 $
- Multiply $ -1 \times -1 = 1 $, add to next coefficient: $ 0 + 1 = 1 $
- Multiply $ 1 \times -1 = -1 $, add to next coefficient: $ 0 + (-1) = -1 $
- Multiply $ -1 \times -1 = 1 $, add to last coefficient: $ -3 + 1 = -2 $

---

### Step 3: Interpret the result

The final row of the synthetic division gives the coefficients of the **quotient** and the **remainder**:

- Quotient: $ x^5 - x^4 + x^3 - x^2 + x - 1 $
- Remainder: $ -2 $

---

### Step 4: Verification

To confirm, we can multiply the divisor $ x + 1 $ by the quotient and add the remainder:

$$
(x + 1)(x^5 - x^4 + x^3 - x^2 + x - 1) + (-2)
$$

Expanding the product:

$$
x^6 - x^5 + x^4 - x^3 + x^2 - x + x^5 - x^4 + x^3 - x^2 + x - 1
$$

Combine like terms:

$$
x^6 + (-x^5 + x^5) + (x^4 - x^4) + (-x^3 + x^3) + (x^2 - x^2) + (-x + x) - 1 = x^6 - 1
$$

Add the remainder $ -2 $:

$$
x^6 - 1 - 2 = x^6 - 3
$$

This matches the original polynomial, confirming the result.

---

### Final Answer

$$
\boxed{x^5 - x^4 + x^3 - x^2 + x - 1}
$$

Solution:
We can perform long division.  Alternatively, by the Remainder Theorem, the remainder upon division is $(-1)^6 - 3 = -2.$  Thus, we can write
\begin{align*}
\frac{x^6 - 3}{x + 1} &= \frac{(x^6 - 1) - 2}{x + 1} \\
&= \frac{x^6 - 1}{x + 1} - \frac{2}{x + 1} \\
&= \frac{(x^3 - 1)(x^3 + 1)}{x + 1} - \frac{2}{x + 1} \\
&= \frac{(x^3 - 1)(x + 1)(x^2 - x + 1)}{x + 1} - \frac{2}{x + 1} \\
&= (x^3 - 1)(x^2 - x + 1) - \frac{2}{x + 1} \\
&= x^5 - x^4 + x^3 - x^2 + x - 1 - \frac{2}{x + 1}.
\end{align*}Thus, the quotient is $\boxed{x^5 - x^4 + x^3 - x^2 + x - 1}.$


Explanation:
Okay, so I need to compute the arcsin of -1/2. Hmm, let me recall what arcsin means. Arcsin is the inverse function of sine, right? So, if I have arcsin(x), it means the angle whose sine is x. But I need to remember the domain and range of arcsin to make sure I get the correct answer.

First, the domain of arcsin is between -1 and 1, which makes sense because the sine function only outputs values in that interval. The range of arcsin is from -π/2 to π/2 radians. Wait, is that right? Let me confirm. Yes, the principal value of arcsin is between -π/2 and π/2. So, the answer should be an angle in that interval whose sine is -1/2.

Alright, so I need to find an angle θ such that sin(θ) = -1/2 and θ is between -π/2 and π/2. Let me think about the unit circle. The sine of an angle is -1/2. I know that sine is negative in the third and fourth quadrants. But since the range of arcsin is restricted to -π/2 to π/2, which is the fourth and first quadrants (since -π/2 is pointing downward along the negative y-axis, and π/2 is upward along the positive y-axis). Wait, actually, between -π/2 and π/2, angles are in the fourth quadrant (negative angles) and first quadrant (positive angles). So, sine is negative in the fourth quadrant. Therefore, the angle we are looking for is in the fourth quadrant.

Now, I need to recall the reference angles where sine is 1/2. The sine of π/6 is 1/2. So, the reference angle is π/6. But since we need a negative sine value, and the angle is in the fourth quadrant, the angle should be -π/6. Let me check that. If θ = -π/6, then sin(-π/6) is equal to -sin(π/6) which is -1/2. That's exactly what we need. So, is that the answer?

Wait, let me make sure there isn't another angle in the range of arcsin that also satisfies sin(θ) = -1/2. The range is from -π/2 to π/2. The other angle where sine is -1/2 is 7π/6, which is in the third quadrant, but 7π/6 is greater than π/2, so it's outside the range of arcsin. Similarly, 11π/6 is in the fourth quadrant, but 11π/6 is equivalent to -π/6 when considering coterminal angles. Wait, 11π/6 is 2π - π/6, which is the same as -π/6. So, yes, -π/6 is the angle in the range of arcsin that corresponds to sine of -1/2.

Let me verify with the unit circle. At -π/6, which is the same as rotating clockwise π/6 radians from the positive x-axis, the coordinates on the unit circle would be (cos(-π/6), sin(-π/6)) which is (cos(π/6), -sin(π/6)) because cosine is even and sine is odd. Cos(π/6) is √3/2 and sin(π/6) is 1/2, so sin(-π/6) is -1/2. Perfect, that matches.

Alternatively, if I think about the sine function's graph, between -π/2 and π/2, it's increasing from -1 to 1. So, the value -1/2 occurs once in that interval, which is at -π/6. Therefore, arcsin(-1/2) is indeed -π/6.

But wait, sometimes people might think of the answer as a positive angle. But since the range of arcsin is restricted to -π/2 to π/2, the answer has to be negative here. If I were to express it as a positive angle, it would be 11π/6, but that's not in the principal value range. So, the correct answer is -π/6.

Let me check with a calculator to be sure. If I calculate arcsin(-1/2), what do I get? Well, I don't have a calculator here, but I can recall that sin(π/6) = 1/2, so sin(-π/6) = -1/2. Therefore, arcsin(-1/2) = -π/6. That seems right.

Alternatively, if I think about the inverse function, since sin(π/6) = 1/2, then arcsin(1/2) = π/6. Similarly, arcsin(-1/2) should be -π/6 because sine is an odd function. So, arcsin(-x) = -arcsin(x). Therefore, arcsin(-1/2) = -arcsin(1/2) = -π/6. That's another way to see it.

So, all these methods point to the answer being -π/6. Therefore, I think that's the correct answer. Let me just make sure there isn't any trick or something I missed. The question says to express the answer in radians, which I have done. The answer is within the principal range of arcsin. So, yes, I think that's solid.

**Final Answer**
The value of $\arcsin \left( -\frac{1}{2} \right)$ is \boxed{-\frac{\pi}{6}}.

To compute $\arcsin\left(-\frac{1}{2}\right)$, we begin by understanding the definition and properties of the inverse sine function.

---

### Step 1: Understand the Domain and Range of $\arcsin$

The function $\arcsin(x)$ is defined for $x \in [-1, 1]$, and its **range** is restricted to:

$$
\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]
$$

This means that the output of $\arcsin(x)$ is always an angle between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$, inclusive. This restriction ensures that $\arcsin$ is a **one-to-one** function, which is necessary for it to be an inverse of the sine function.

---

### Step 2: Identify the Reference Angle

We are looking for an angle $\theta$ such that:

$$
\sin(\theta) = -\frac{1}{2}
$$

We know that:

$$
\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}
$$

So, the **reference angle** is $\frac{\pi}{6}$. Since the sine value is **negative**, the angle must lie in a quadrant where sine is negative. Within the range of $\arcsin$, which is between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$, the only quadrant where sine is negative is the **fourth quadrant**.

---

### Step 3: Determine the Correct Angle

In the fourth quadrant, the angle corresponding to a reference angle of $\frac{\pi}{6}$ is:

$$
-\frac{\pi}{6}
$$

This is because:

$$
\sin\left(-\frac{\pi}{6}\right) = -\sin\left(\frac{\pi}{6}\right) = -\frac{1}{2}
$$

And $-\frac{\pi}{6}$ lies within the range of $\arcsin$, so it is the correct value.

---

### Step 4: Final Answer

Thus, the value of:

$$
\arcsin\left(-\frac{1}{2}\right) = \boxed{-\frac{\pi}{6}}
$$

[Assistant]
Explanation:
Okay, so I need to compute the arcsin of -1/2. Hmm, let me recall what arcsin means. Arcsin is the inverse function of sine, right? So, if I have arcsin(x), it means the angle whose sine is x. But I need to remember the domain and range of arcsin to make sure I get the correct answer.

First, the domain of arcsin is between -1 and 1, which makes sense because the sine function only outputs values in that interval. The range of arcsin is from -π/2 to π/2 radians. Wait, is that right? Let me confirm. Yes, the principal value of arcsin is between -π/2 and π/2. So, the answer should be an angle in that interval whose sine is -1/2.

Alright, so I need to find an angle θ such that sin(θ) = -1/2 and θ is between -π/2 and π/2. Let me think about the unit circle. The sine of an angle is -1/2. I know that sine is negative in the third and fourth quadrants. But since the range of arcsin is restricted to -π/2 to π/2, which is the fourth and first quadrants (since -π/2 is pointing downward along the negative y-axis, and π/2 is upward along the positive y-axis). Wait, actually, between -π/2 and π/2, angles are in the fourth quadrant (negative angles) and first quadrant (positive angles). So, sine is negative in the fourth quadrant. Therefore, the angle we are looking for is in the fourth quadrant.

Now, I need to recall the reference angles where sine is 1/2. The sine of π/6 is 1/2. So, the reference angle is π/6. But since we need a negative sine value, and the angle is in the fourth quadrant, the angle should be -π/6. Let me check that. If θ = -π/6, then sin(-π/6) is equal to -sin(π/6) which is -1/2. That's exactly what we need. So, is that the answer?

Wait, let me make sure there isn't another angle in the range of arcsin that also satisfies sin(θ) = -1/2. The range is from -π/2 to π/2. The other angle where sine is -1/2 is 7π/6, which is in the third quadrant, but 7π/6 is greater than π/2, so it's outside the range of arcsin. Similarly, 11π/6 is in the fourth quadrant, but 11π/6 is equivalent to -π/6 when considering coterminal angles. Wait, 11π/6 is 2π - π/6, which is the same as -π/6. So, yes, -π/6 is the angle in the range of arcsin that corresponds to sine of -1/2.

Let me verify with the unit circle. At -π/6, which is the same as rotating clockwise π/6 radians from the positive x-axis, the coordinates on the unit circle would be (cos(-π/6), sin(-π/6)) which is (cos(π/6), -sin(π/6)) because cosine is even and sine is odd. Cos(π/6) is √3/2 and sin(π/6) is 1/2, so sin(-π/6) is -1/2. Perfect, that matches.

Alternatively, if I think about the sine function's graph, between -π/2 and π/2, it's increasing from -1 to 1. So, the value -1/2 occurs once in that interval, which is at -π/6. Therefore, arcsin(-1/2) is indeed -π/6.

But wait, sometimes people might think of the answer as a positive angle. But since the range of arcsin is restricted to -π/2 to π/2, the answer has to be negative here. If I were to express it as a positive angle, it would be 11π/6, but that's not in the principal value range. So, the correct answer is -π/6.

Let me check with a calculator to be sure. If I calculate arcsin(-1/2), what do I get? Well, I don't have a calculator here, but I can recall that sin(π/6) = 1/2, so sin(-π/6) = -1/2. Therefore, arcsin(-1/2) = -π/6. That seems right.

Alternatively, if I think about the inverse function, since sin(π/6) = 1/2, then arcsin(1/2) = π/6. Similarly, arcsin(-1/2) should be -π/6 because sine is an odd function. So, arcsin(-x) = -arcsin(x). Therefore, arcsin(-1/2) = -arcsin(1/2) = -π/6. That's another way to see it.

So, all these methods point to the answer being -π/6. Therefore, I think that's the correct answer. Let me just make sure there isn't any trick or something I missed. The question says to express the answer in radians, which I have done. The answer is within the principal range of arcsin. So, yes, I think that's solid.

**Final Answer**
The value of $\arcsin \left( -\frac{1}{2} \right)$ is \boxed{-\frac{\pi}{6}}.

To compute $\arcsin\left(-\frac{1}{2}\right)$, we begin by understanding the definition and properties of the inverse sine function.

---

### Step 1: Understand the Domain and Range of $\arcsin$

The function $\arcsin(x)$ is defined for $x \in [-1, 1]$, and its **range** is restricted to:

$$
\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]
$$

This means that the output of $\arcsin(x)$ is always an angle between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$, inclusive. This restriction ensures that $\arcsin$ is a **one-to-one** function, which is necessary for it to be an inverse of the sine function.

---

### Step 2: Identify the Reference Angle

We are looking for an angle $\theta$ such that:

$$
\sin(\theta) = -\frac{1}{2}
$$

We know that:

$$
\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}
$$

So, the **reference angle** is $\frac{\pi}{6}$. Since the sine value is **negative**, the angle must lie in a quadrant where sine is negative. Within the range of $\arcsin$, which is between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$, the only quadrant where sine is negative is the **fourth quadrant**.

---

### Step 3: Determine the Correct Angle

In the fourth quadrant, the angle corresponding to a reference angle of $\frac{\pi}{6}$ is:

$$
-\frac{\pi}{6}
$$

This is because:

$$
\sin\left(-\frac{\pi}{6}\right) = -\sin\left(\frac{\pi}{6}\right) = -\frac{1}{2}
$$

And $-\frac{\pi}{6}$ lies within the range of $\arcsin$, so it is the correct value.

---

### Step 4: Final Answer

Thus, the value of:

$$
\arcsin\left(-\frac{1}{2}\right) = \boxed{-\frac{\pi}{6}}
$$

Solution: Since $\sin \left( -\frac{\pi}{6} \right) = -\frac{1}{2},$ $\arcsin \left( -\frac{1}{2} \right) = \boxed{-\frac{\pi}{6}}.$


Explanation:
Okay, so I need to convert the fraction 3/20 into a decimal. Hmm, let me think about how to do that. I remember that converting a fraction to a decimal involves dividing the numerator by the denominator. So in this case, I need to divide 3 by 20.

Wait, let me make sure I remember that correctly. Yeah, fractions are just another way of writing division, right? So 3/20 is the same as 3 divided by 20. Alright, so I need to perform that division. But how exactly do I do that? Let me recall.

I think there are a couple of ways to do this. One way is to use long division, where I divide 3 by 20. But since 20 is larger than 3, the result will be less than 1. Maybe I can add a decimal point and some zeros to make the division work. Let me try that.

So, if I write 3 divided by 20, I can think of it as 3.000 divided by 20. Let me set that up.

First, 20 goes into 3 zero times. So I put a 0. and then proceed. Then, 20 goes into 30 once, right? Because 20 times 1 is 20. Subtract 20 from 30, and I get 10. Then bring down the next 0, making it 100.

Wait, hold on. Let me write this step by step.

Starting with 3 divided by 20:

- 20 | 3.000

Since 20 doesn't go into 3, we write 0. and then consider 30 (by adding a decimal point and a zero).

So, 20 goes into 30 once (1). Multiply 1 by 20, subtract from 30, we get 10. Bring down the next 0, making it 100.

Now, how many times does 20 go into 100? Let me calculate. 20 times 5 is 100. So that's 5. Multiply 5 by 20, subtract from 100, we get 0.

So, putting that together, we have 0.15. Wait, so 3 divided by 20 is 0.15? Let me check that.

Alternatively, maybe I can think of 3/20 as a fraction that can be converted to a decimal by multiplying numerator and denominator by something to make the denominator a power of 10. Let me try that method.

The denominator is 20. If I can make it 100, that would be helpful because then the decimal would be straightforward.

So, 20 times 5 is 100. Therefore, if I multiply numerator and denominator by 5, I get:

(3 * 5) / (20 * 5) = 15 / 100.

And 15/100 is 0.15.

Oh, that's a quicker way. So that confirms the previous result.

Alternatively, maybe I can think of 3/20 as 3 divided by 2 divided by 10. Let me see.

First, divide 3 by 2, which is 1.5, and then divide by 10, which would be 0.15.

Yes, that also gives me 0.15.

So all three methods give me the same answer. That seems consistent.

But let me verify once more with another approach. Maybe using decimal equivalents of fractions I know.

I know that 1/20 is 0.05, right? Because 1 divided by 20 is 0.05. So if I multiply that by 3, I get 0.05 * 3 = 0.15.

Yep, that's another way to see it.

So, all these different methods lead me to the conclusion that 3/20 as a decimal is 0.15.

But just to be thorough, let me do the long division again step by step.

Dividing 3 by 20:

- 20 goes into 3 zero times. Write 0.
- Add a decimal point and a zero, making it 30.
- 20 goes into 30 once (1). Multiply 1 by 20, subtract from 30, remainder 10.
- Bring down another zero, making it 100.
- 20 goes into 100 five times (5). Multiply 5 by 20, subtract from 100, remainder 0.
- No more remainders, so the division ends here.

So the result is 0.15.

Therefore, I can be confident that 3/20 is equal to 0.15.

Alternatively, if I use a calculator to check, 3 divided by 20 equals... let me imagine pressing the buttons. 3 ÷ 20 = 0.15. Yep, that's correct.

So, all methods converge to 0.15.

I think that's solid. I don't see any mistakes in my reasoning. Each step checks out.

Another way to think about it: 20 times 0.1 is 2, and 20 times 0.05 is 1. So 0.1 + 0.05 = 0.15, and 20 times 0.15 is 3. Exactly.

Therefore, 3/20 is indeed 0.15.

I think that's thorough enough. I can't think of any other way to approach this problem, and all methods confirm the same answer.

**Final Answer**
The decimal form of $\frac{3}{20}$ is \boxed{0.15}.

To convert the fraction $\frac{3}{20}$ into a decimal, we can use the fundamental concept that a fraction represents division. That is, $\frac{3}{20}$ is equivalent to $3 \div 20$.

---

### Step-by-Step Conversion Using Long Division

1. **Set up the division**:
   We divide 3 by 20. Since 20 is larger than 3, the result will be less than 1.
   We write this as $3.000 \div 20$.

2. **First step**:
   - 20 goes into 3 zero times.
   - Place a decimal point after the 0 and bring down a zero, making it 30.

3. **Second step**:
   - 20 goes into 30 once (1).
   - Multiply 1 by 20, subtract from 30, remainder is 10.

4. **Third step**:
   - Bring down another zero, making it 100.
   - 20 goes into 100 five times (5).
   - Multiply 5 by 20, subtract from 100, remainder is 0.

5. **Final result**:
   The division ends here, and the quotient is **0.15**.

---

### Alternative Method: Converting to a Decimal by Adjusting the Denominator

We can also convert the fraction by adjusting the denominator to a power of 10:

- Multiply both numerator and denominator by 5:
  $$
  \frac{3}{20} = \frac{3 \times 5}{20 \times 5} = \frac{15}{100}
  $$

- $\frac{15}{100}$ is equivalent to **0.15**.

---

### Verification Using Known Decimal Equivalents

- We know that $\frac{1}{20} = 0.05$.
- Therefore, $\frac{3}{20} = 3 \times 0.05 = 0.15$.

---

### Final Answer

$$
\boxed{0.15}
$$

[Assistant]
Explanation:
Okay, so I need to convert the fraction 3/20 into a decimal. Hmm, let me think about how to do that. I remember that converting a fraction to a decimal involves dividing the numerator by the denominator. So in this case, I need to divide 3 by 20.

Wait, let me make sure I remember that correctly. Yeah, fractions are just another way of writing division, right? So 3/20 is the same as 3 divided by 20. Alright, so I need to perform that division. But how exactly do I do that? Let me recall.

I think there are a couple of ways to do this. One way is to use long division, where I divide 3 by 20. But since 20 is larger than 3, the result will be less than 1. Maybe I can add a decimal point and some zeros to make the division work. Let me try that.

So, if I write 3 divided by 20, I can think of it as 3.000 divided by 20. Let me set that up.

First, 20 goes into 3 zero times. So I put a 0. and then proceed. Then, 20 goes into 30 once, right? Because 20 times 1 is 20. Subtract 20 from 30, and I get 10. Then bring down the next 0, making it 100.

Wait, hold on. Let me write this step by step.

Starting with 3 divided by 20:

- 20 | 3.000

Since 20 doesn't go into 3, we write 0. and then consider 30 (by adding a decimal point and a zero).

So, 20 goes into 30 once (1). Multiply 1 by 20, subtract from 30, we get 10. Bring down the next 0, making it 100.

Now, how many times does 20 go into 100? Let me calculate. 20 times 5 is 100. So that's 5. Multiply 5 by 20, subtract from 100, we get 0.

So, putting that together, we have 0.15. Wait, so 3 divided by 20 is 0.15? Let me check that.

Alternatively, maybe I can think of 3/20 as a fraction that can be converted to a decimal by multiplying numerator and denominator by something to make the denominator a power of 10. Let me try that method.

The denominator is 20. If I can make it 100, that would be helpful because then the decimal would be straightforward.

So, 20 times 5 is 100. Therefore, if I multiply numerator and denominator by 5, I get:

(3 * 5) / (20 * 5) = 15 / 100.

And 15/100 is 0.15.

Oh, that's a quicker way. So that confirms the previous result.

Alternatively, maybe I can think of 3/20 as 3 divided by 2 divided by 10. Let me see.

First, divide 3 by 2, which is 1.5, and then divide by 10, which would be 0.15.

Yes, that also gives me 0.15.

So all three methods give me the same answer. That seems consistent.

But let me verify once more with another approach. Maybe using decimal equivalents of fractions I know.

I know that 1/20 is 0.05, right? Because 1 divided by 20 is 0.05. So if I multiply that by 3, I get 0.05 * 3 = 0.15.

Yep, that's another way to see it.

So, all these different methods lead me to the conclusion that 3/20 as a decimal is 0.15.

But just to be thorough, let me do the long division again step by step.

Dividing 3 by 20:

- 20 goes into 3 zero times. Write 0.
- Add a decimal point and a zero, making it 30.
- 20 goes into 30 once (1). Multiply 1 by 20, subtract from 30, remainder 10.
- Bring down another zero, making it 100.
- 20 goes into 100 five times (5). Multiply 5 by 20, subtract from 100, remainder 0.
- No more remainders, so the division ends here.

So the result is 0.15.

Therefore, I can be confident that 3/20 is equal to 0.15.

Alternatively, if I use a calculator to check, 3 divided by 20 equals... let me imagine pressing the buttons. 3 ÷ 20 = 0.15. Yep, that's correct.

So, all methods converge to 0.15.

I think that's solid. I don't see any mistakes in my reasoning. Each step checks out.

Another way to think about it: 20 times 0.1 is 2, and 20 times 0.05 is 1. So 0.1 + 0.05 = 0.15, and 20 times 0.15 is 3. Exactly.

Therefore, 3/20 is indeed 0.15.

I think that's thorough enough. I can't think of any other way to approach this problem, and all methods confirm the same answer.

**Final Answer**
The decimal form of $\frac{3}{20}$ is \boxed{0.15}.

To convert the fraction $\frac{3}{20}$ into a decimal, we can use the fundamental concept that a fraction represents division. That is, $\frac{3}{20}$ is equivalent to $3 \div 20$.

---

### Step-by-Step Conversion Using Long Division

1. **Set up the division**:
   We divide 3 by 20. Since 20 is larger than 3, the result will be less than 1.
   We write this as $3.000 \div 20$.

2. **First step**:
   - 20 goes into 3 zero times.
   - Place a decimal point after the 0 and bring down a zero, making it 30.

3. **Second step**:
   - 20 goes into 30 once (1).
   - Multiply 1 by 20, subtract from 30, remainder is 10.

4. **Third step**:
   - Bring down another zero, making it 100.
   - 20 goes into 100 five times (5).
   - Multiply 5 by 20, subtract from 100, remainder is 0.

5. **Final result**:
   The division ends here, and the quotient is **0.15**.

---

### Alternative Method: Converting to a Decimal by Adjusting the Denominator

We can also convert the fraction by adjusting the denominator to a power of 10:

- Multiply both numerator and denominator by 5:
  $$
  \frac{3}{20} = \frac{3 \times 5}{20 \times 5} = \frac{15}{100}
  $$

- $\frac{15}{100}$ is equivalent to **0.15**.

---

### Verification Using Known Decimal Equivalents

- We know that $\frac{1}{20} = 0.05$.
- Therefore, $\frac{3}{20} = 3 \times 0.05 = 0.15$.

---

### Final Answer

$$
\boxed{0.15}
$$

Solution: Multiplying numerator and denominator by 5 gives $3/20=15/100 = \boxed{0.15}$.


Explanation:
Okay, so Ringo needs to calculate the modulus of (1 - i) raised to the 8th power. Hmm, let me think. First, I remember that the modulus of a complex number a + bi is sqrt(a² + b²). So maybe I should start by finding the modulus of (1 - i) first, and then raise it to the 8th power? Or maybe there's another way? Wait, I also recall that when you raise a complex number to a power, you can use De Moivre's theorem, which involves the modulus and the argument. Let me try both methods and see if they match.

First, let me compute |1 - i|. The modulus is sqrt(1² + (-1)²) = sqrt(1 + 1) = sqrt(2). So |1 - i| is sqrt(2). Then, if I raise that modulus to the 8th power, it would be (sqrt(2))⁸. Let me calculate that. Since sqrt(2) is 2^(1/2), raising it to the 8th power would be 2^(1/2 * 8) = 2^4 = 16. So that would be 16. Is that right? Let me check with another method.

Alternatively, using De Moivre's theorem. First, express 1 - i in polar form. The modulus is sqrt(2) as before. The argument θ is the angle made with the positive real axis. Since the complex number is 1 - i, it's in the fourth quadrant. The reference angle would be arctangent of |b/a| = arctan(1/1) = π/4. So the argument θ is -π/4 or 7π/4. So in polar form, 1 - i is sqrt(2)*(cos(-π/4) + i sin(-π/4)).

Now, raising this to the 8th power: [sqrt(2)]^8 * [cos(8*(-π/4)) + i sin(8*(-π/4))]. Wait, but the modulus part is [sqrt(2)]^8 which we already calculated as 16. Then the angle part is 8*(-π/4) = -2π. But cosine and sine are periodic with period 2π, so cos(-2π) = cos(0) = 1, and sin(-2π) = sin(0) = 0. Therefore, the result is 16*(1 + 0i) = 16. So that matches the previous result. So the modulus is 16. Wait, but the question is asking for |(1 - i)^8|. So modulus of (1 - i)^8 is equal to |1 - i|^8, which is (sqrt(2))^8 = 16. So that's straightforward.

Alternatively, maybe I can compute (1 - i)^8 directly? Let me try that. Although that might be more work, but let's see. Let me compute (1 - i)^2 first. (1 - i)^2 = 1^2 - 2*1*i + i^2 = 1 - 2i + (-1) = 0 - 2i = -2i. Then, (1 - i)^4 is [(1 - i)^2]^2 = (-2i)^2 = 4i^2 = 4*(-1) = -4. Then, (1 - i)^8 is [(1 - i)^4]^2 = (-4)^2 = 16. Oh, that's even simpler! So that also gives 16. So all methods lead to 16. Therefore, Ringo should have found 16.

But let me just make sure I didn't make any mistakes in the calculations. Let me check each step again.

First method: modulus of (1 - i) is sqrt(1^2 + (-1)^2) = sqrt(2). Then modulus of (1 - i)^8 is |1 - i|^8 = (sqrt(2))^8. Since sqrt(2) is 2^(1/2), raising to the 8th power gives 2^(4) = 16. Correct.

Second method: polar form. 1 - i has modulus sqrt(2) and angle -π/4. Raising to 8th power: modulus becomes (sqrt(2))^8 = 16, angle becomes 8*(-π/4) = -2π, which is equivalent to 0 radians. So the complex number is 16*(cos 0 + i sin 0) = 16*1 + 16*0i = 16. Correct.

Third method: expanding step by step. (1 - i)^2 = -2i. Then (1 - i)^4 = (-2i)^2 = -4. Then (1 - i)^8 = (-4)^2 = 16. Correct. All methods confirm that the answer is 16. Therefore, Ringo should have found 16. I think that's solid. No mistakes detected here.

**Final Answer**
\boxed{16}

To solve the expression $ |(1 - i)^8| $, we can approach the problem using properties of complex numbers and their modulus.

---

### Step 1: Understanding the Modulus of a Complex Number

For any complex number $ z = a + bi $, the modulus is defined as:

$$
|z| = \sqrt{a^2 + b^2}
$$

In this case, the complex number is $ 1 - i $, so:

$$
|1 - i| = \sqrt{1^2 + (-1)^2} = \sqrt{1 + 1} = \sqrt{2}
$$

---

### Step 2: Using the Property of Modulus with Powers

A key property of modulus is that for any complex number $ z $ and integer $ n $:

$$
|z^n| = |z|^n
$$

So, applying this to our problem:

$$
|(1 - i)^8| = |1 - i|^8 = (\sqrt{2})^8
$$

Now, simplify $ (\sqrt{2})^8 $:

$$
(\sqrt{2})^8 = (2^{1/2})^8 = 2^{(1/2) \cdot 8} = 2^4 = 16
$$

---

### Step 3: Verification via Direct Expansion (Optional)

As a cross-check, we can compute $ (1 - i)^8 $ directly:

- $ (1 - i)^2 = 1 - 2i + i^2 = 1 - 2i - 1 = -2i $
- $ (1 - i)^4 = (-2i)^2 = 4i^2 = -4 $
- $ (1 - i)^8 = (-4)^2 = 16 $

This confirms the result.

---

### Final Answer

$$
\boxed{16}
$$

[Assistant]
Explanation:
Okay, so Ringo needs to calculate the modulus of (1 - i) raised to the 8th power. Hmm, let me think. First, I remember that the modulus of a complex number a + bi is sqrt(a² + b²). So maybe I should start by finding the modulus of (1 - i) first, and then raise it to the 8th power? Or maybe there's another way? Wait, I also recall that when you raise a complex number to a power, you can use De Moivre's theorem, which involves the modulus and the argument. Let me try both methods and see if they match.

First, let me compute |1 - i|. The modulus is sqrt(1² + (-1)²) = sqrt(1 + 1) = sqrt(2). So |1 - i| is sqrt(2). Then, if I raise that modulus to the 8th power, it would be (sqrt(2))⁸. Let me calculate that. Since sqrt(2) is 2^(1/2), raising it to the 8th power would be 2^(1/2 * 8) = 2^4 = 16. So that would be 16. Is that right? Let me check with another method.

Alternatively, using De Moivre's theorem. First, express 1 - i in polar form. The modulus is sqrt(2) as before. The argument θ is the angle made with the positive real axis. Since the complex number is 1 - i, it's in the fourth quadrant. The reference angle would be arctangent of |b/a| = arctan(1/1) = π/4. So the argument θ is -π/4 or 7π/4. So in polar form, 1 - i is sqrt(2)*(cos(-π/4) + i sin(-π/4)).

Now, raising this to the 8th power: [sqrt(2)]^8 * [cos(8*(-π/4)) + i sin(8*(-π/4))]. Wait, but the modulus part is [sqrt(2)]^8 which we already calculated as 16. Then the angle part is 8*(-π/4) = -2π. But cosine and sine are periodic with period 2π, so cos(-2π) = cos(0) = 1, and sin(-2π) = sin(0) = 0. Therefore, the result is 16*(1 + 0i) = 16. So that matches the previous result. So the modulus is 16. Wait, but the question is asking for |(1 - i)^8|. So modulus of (1 - i)^8 is equal to |1 - i|^8, which is (sqrt(2))^8 = 16. So that's straightforward.

Alternatively, maybe I can compute (1 - i)^8 directly? Let me try that. Although that might be more work, but let's see. Let me compute (1 - i)^2 first. (1 - i)^2 = 1^2 - 2*1*i + i^2 = 1 - 2i + (-1) = 0 - 2i = -2i. Then, (1 - i)^4 is [(1 - i)^2]^2 = (-2i)^2 = 4i^2 = 4*(-1) = -4. Then, (1 - i)^8 is [(1 - i)^4]^2 = (-4)^2 = 16. Oh, that's even simpler! So that also gives 16. So all methods lead to 16. Therefore, Ringo should have found 16.

But let me just make sure I didn't make any mistakes in the calculations. Let me check each step again.

First method: modulus of (1 - i) is sqrt(1^2 + (-1)^2) = sqrt(2). Then modulus of (1 - i)^8 is |1 - i|^8 = (sqrt(2))^8. Since sqrt(2) is 2^(1/2), raising to the 8th power gives 2^(4) = 16. Correct.

Second method: polar form. 1 - i has modulus sqrt(2) and angle -π/4. Raising to 8th power: modulus becomes (sqrt(2))^8 = 16, angle becomes 8*(-π/4) = -2π, which is equivalent to 0 radians. So the complex number is 16*(cos 0 + i sin 0) = 16*1 + 16*0i = 16. Correct.

Third method: expanding step by step. (1 - i)^2 = -2i. Then (1 - i)^4 = (-2i)^2 = -4. Then (1 - i)^8 = (-4)^2 = 16. Correct. All methods confirm that the answer is 16. Therefore, Ringo should have found 16. I think that's solid. No mistakes detected here.

**Final Answer**
\boxed{16}

To solve the expression $ |(1 - i)^8| $, we can approach the problem using properties of complex numbers and their modulus.

---

### Step 1: Understanding the Modulus of a Complex Number

For any complex number $ z = a + bi $, the modulus is defined as:

$$
|z| = \sqrt{a^2 + b^2}
$$

In this case, the complex number is $ 1 - i $, so:

$$
|1 - i| = \sqrt{1^2 + (-1)^2} = \sqrt{1 + 1} = \sqrt{2}
$$

---

### Step 2: Using the Property of Modulus with Powers

A key property of modulus is that for any complex number $ z $ and integer $ n $:

$$
|z^n| = |z|^n
$$

So, applying this to our problem:

$$
|(1 - i)^8| = |1 - i|^8 = (\sqrt{2})^8
$$

Now, simplify $ (\sqrt{2})^8 $:

$$
(\sqrt{2})^8 = (2^{1/2})^8 = 2^{(1/2) \cdot 8} = 2^4 = 16
$$

---

### Step 3: Verification via Direct Expansion (Optional)

As a cross-check, we can compute $ (1 - i)^8 $ directly:

- $ (1 - i)^2 = 1 - 2i + i^2 = 1 - 2i - 1 = -2i $
- $ (1 - i)^4 = (-2i)^2 = 4i^2 = -4 $
- $ (1 - i)^8 = (-4)^2 = 16 $

This confirms the result.

---

### Final Answer

$$
\boxed{16}
$$

Solution: We know that magnitudes of complex numbers are multiplicative: the magnitude of $|ab|$ is the product $|a|\cdot |b|$. Thus, \[\left|\left(1-i\right)^8\right|=\left|1-i\right|^8\]The magnitude of $1-i$ is $\sqrt{1^2+(-1)^2}=\sqrt{2}$; thus our answer is $\left(\sqrt{2}\right) ^8=\boxed{16}$. Ringo didn't give much of a reprieve.