Okay, let's see. I need to find the absolute value of (1 - i)^8. Hmm. Alright, first, I remember that when dealing with complex numbers, the absolute value or modulus of a complex number z = a + bi is sqrt(a² + b²). But here, it's a complex number raised to the 8th power. So maybe I need to use some properties of complex numbers here.

Wait, the modulus of a product of complex numbers is the product of their moduli. So |z^n| = |z|^n. Is that right? Let me confirm. If z and w are complex numbers, then |zw| = |z||w|. So by induction, |z^n| = |z|^n. Yeah, that seems correct. So maybe I can just compute the modulus of (1 - i) first and then raise it to the 8th power?

Let me try that. The modulus of (1 - i) is sqrt(1² + (-1)²) = sqrt(1 + 1) = sqrt(2). So |1 - i| = sqrt(2). Then |(1 - i)^8| = (sqrt(2))^8. Let's compute that. (sqrt(2))^8 is the same as (2^(1/2))^8 = 2^(8/2) = 2^4 = 16. So is the answer 16? That seems straightforward.

But wait, maybe I should check another way to make sure I didn't make a mistake. Let's compute (1 - i)^8 step by step. Maybe expanding it using binomial theorem would help. Let's see.

First, compute (1 - i)^2. That's (1)^2 + (-i)^2 + 2*(1)*(-i) = 1 + (-1) + (-2i) = 0 - 2i = -2i. Wait, that's interesting. So (1 - i)^2 = -2i. Then, (1 - i)^4 would be [(-2i)]^2 = (-2)^2 * (i)^2 = 4 * (-1) = -4. Then (1 - i)^8 would be (-4)^2 = 16. So that's the same result. So that confirms it. So the modulus is 16.

Alternatively, maybe using polar form. Let's express (1 - i) in polar form. The modulus is sqrt(2), as before. The argument is arctan(-1/1) = -π/4. So in polar form, (1 - i) = sqrt(2) * (cos(-π/4) + i sin(-π/4)). Then raising it to the 8th power: [sqrt(2)]^8 * [cos(8*(-π/4)) + i sin(8*(-π/4))]. Let's compute that. [sqrt(2)]^8 is (2^(1/2))^8 = 2^(4) = 16. The angle is 8*(-π/4) = -2π. Cos(-2π) = cos(2π) = 1, sin(-2π) = sin(2π) = 0. So the result is 16*(1 + 0i) = 16. So again, modulus is 16. So all methods lead to the same answer. Therefore, the answer should be 16.

Wait, but the problem says "|(1 - i)^8|". Since we're taking the modulus of a complex number, which is a real number, the answer is just 16. So yeah, I think that's correct. Let me check once more with another approach. Maybe using the binomial expansion for (1 - i)^8. Although that might be tedious, but let's see.

(1 - i)^1 = 1 - i

(1 - i)^2 = (1 - i)(1 - i) = 1 - 2i + i² = 1 - 2i -1 = -2i

(1 - i)^3 = (1 - i)^2*(1 - i) = (-2i)(1 - i) = -2i + 2i² = -2i -2 = -2 - 2i

(1 - i)^4 = (1 - i)^3*(1 - i) = (-2 - 2i)(1 - i) = -2(1 - i) -2i(1 - i) = -2 + 2i -2i + 2i² = -2 + 0i -2 = -4

(1 - i)^5 = (1 - i)^4*(1 - i) = (-4)(1 - i) = -4 + 4i

(1 - i)^6 = (1 - i)^5*(1 - i) = (-4 + 4i)(1 - i) = -4(1 - i) + 4i(1 - i) = -4 + 4i +4i -4i² = -4 +8i +4 = 8i

(1 - i)^7 = (1 - i)^6*(1 - i) = 8i*(1 - i) = 8i -8i² = 8i +8 = 8 +8i

(1 - i)^8 = (1 - i)^7*(1 - i) = (8 +8i)(1 - i) = 8(1 - i) +8i(1 - i) = 8 -8i +8i -8i² = 8 +0i +8 = 16

So expanding step by step, we also get 16. Therefore, the modulus is indeed 16. So all methods confirm that the answer is 16. Therefore, Ringo should have found 16.
Therefore, the absolute value of \((1 - i)^8\) is \(\boxed{16}\).
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