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Home » Since 125 is not a power of 2, continue splitting down to 1: - Leaselab

Since 125 is not a power of 2, continue splitting down to 1: - Leaselab

📅 February 27, 2026 👤 scraface

Feb 27, 2026

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📌 Let’s find the roots of \( z^3 = \omega \). Write \( \omega = e^{2\pi i/3} \), so:

📌 z = \omega^{1/3} \cdot e^{2\pi i k / 3} = e^{2\pi i/9} \cdot e^{2\pi i k / 3} = e^{2\pi i (1/9 + k/3)} = e^{2\pi i (1/9 + 3k/9)} = e^{2\pi i (3k+1)/9}, \quad k = 0,1,2

📌 Similarly, for \( z^3 = \omega^2 = e^{-2\pi i/3} \):

📌 z = e^{-2\pi i/9} \cdot e^{2\pi i k / 3} = e^{-2\pi i (1 - 3k)/9} = e^{2\pi i (2/9 + k/3)} = e^{2\pi i (2 + 3k)/9}, \quad k = 0,1,2

📌 So all roots have arguments:

📌 From first set: \( \theta = \frac{2\pi(3k+1)}{9} \), \( k=0,1,2 \) → \( \frac{2\pi}{9}, \frac{8\pi}{9}, \frac{14\pi}{9} \)

📌 From second set: \( \theta = \frac{2\pi(3k+2)}{9} \), \( k=0,1,2 \) → \( \frac{4\pi}{9}, \frac{10\pi}{9}, \frac{16\pi}{9} \)

📌 List the real parts: \( \cos\theta \) for each:

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= 3 - 6i + 4i - 8i^2
Recall that \(i^2 = -1\). Substituting, we have:
= 3 - 6i + 4i + 8
= (3 + 8) + (-6i + 4i)
= 11 - 2i
Thus, the product of the force fields is:
\boxed{11 - 2i}
Question:** A philosopher of science contemplates the growth of scient...
We will compute the terms of the sequence step-by-step using the recur...
\(a_1 = 1\)

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