QuestionAugust 25, 2025

Look at this equation: v^3=1 What is v, the cube root of 1? V= square

Look at this equation: v^3=1 What is v, the cube root of 1? V= square
Look at this equation: v^3=1 What is v, the cube root of 1? V= square

Solution
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Answer

V = 1, -\frac{1}{2} + \frac{\sqrt{3}}{2}i, -\frac{1}{2} - \frac{\sqrt{3}}{2}i Explanation 1. Identify the roots of unity The equation v^3 = 1 implies v is a cube root of unity. The cube roots of unity are given by the formula v = e^{2\pi i k / 3} for k = 0, 1, 2. 2. Calculate the roots For k = 0: v = e^{2\pi i \cdot 0 / 3} = 1. For k = 1: v = e^{2\pi i \cdot 1 / 3} = e^{2\pi i / 3} = -\frac{1}{2} + \frac{\sqrt{3}}{2}i. For k = 2: v = e^{2\pi i \cdot 2 / 3} = e^{4\pi i / 3} = -\frac{1}{2} - \frac{\sqrt{3}}{2}i.

Explanation

1. Identify the roots of unity<br /> The equation $v^3 = 1$ implies $v$ is a cube root of unity. The cube roots of unity are given by the formula $v = e^{2\pi i k / 3}$ for $k = 0, 1, 2$.<br /><br />2. Calculate the roots<br /> For $k = 0$: $v = e^{2\pi i \cdot 0 / 3} = 1$.<br /> For $k = 1$: $v = e^{2\pi i \cdot 1 / 3} = e^{2\pi i / 3} = -\frac{1}{2} + \frac{\sqrt{3}}{2}i$.<br /> For $k = 2$: $v = e^{2\pi i \cdot 2 / 3} = e^{4\pi i / 3} = -\frac{1}{2} - \frac{\sqrt{3}}{2}i$.
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