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Chapter 1: Problem 2

Show that \\[ \frac{-1+\sqrt{3} i}{2} \\] is a cube root of 1 (meaning that its cube equals 1 ).

Short Answer

Expert verified
Computing the cube of the given complex number \((\frac{-1+\sqrt{3}i}{2})^3\) and simplifying the result, we find that it equals 1. Therefore, \(\frac{-1+\sqrt{3}i}{2}\) is a cube root of 1.

Step by step solution

01

Write down the complex number we want to check

We are given the complex number \(\frac{-1+\sqrt{3}i}{2}\), and we want to compute its cube to check if it is a cube root of 1.
02

Compute the cube of the complex number

To compute the cube of the complex number \((\frac{-1+\sqrt{3}i}{2})^3\), we multiply it by itself three times: \[(\frac{-1+\sqrt{3}i}{2})^3 = (\frac{-1+\sqrt{3}i}{2})(\frac{-1+\sqrt{3}i}{2})(\frac{-1+\sqrt{3}i}{2}).\]
03

Multiply the complex numbers

Now we multiply the complex numbers in pairs: \[(\frac{-1+\sqrt{3}i}{2})(\frac{-1+\sqrt{3}i}{2}) = \frac{(-1)(-1)+(-1)(\sqrt{3}i)+(\sqrt{3}i)(-1)+(\sqrt{3}i)(\sqrt{3}i)}{2\cdot2} = \frac{1 - \sqrt{3}i - \sqrt{3}i + 3i^2}{4}\]
04

Simplify the result

Recall that \(i^2 = -1\), so we simplify the last expression: \[\frac{1 - \sqrt{3}i - \sqrt{3}i - 3}{4} = \frac{-2 - 2\sqrt{3}i}{4}\]
05

Multiply the simplified result with the remaining complex number

Now we multiply the simplified result with the remaining complex number: \[\frac{-2 - 2\sqrt{3}i}{4} \cdot \frac{-1+\sqrt{3}i}{2} = \frac{(-2)(-1) + (-2)(\sqrt{3}i) - 2\sqrt{3}i(\sqrt{3}i) - 2\sqrt{3}i}{4\cdot2}\]
06

Simplify the last expression

Notice that we can simplify the expression further, using the algebraic properties of complex numbers: \[\frac{2 - 2\sqrt{3}i + 6 - 2\sqrt{3}i}{8} = \frac{8}{8} = 1\] Since the cube of \(\frac{-1+\sqrt{3}i}{2}\) equals 1, we can conclude that \(\frac{-1+\sqrt{3}i}{2}\) is a cube root of 1.

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Most popular questions from this chapter

Suppose \(a\) and \(b\) are real numbers, not both 0. Find real numbers \(c\) and \(d\) such that \\[ 1 /(a+b i)=c+d i \\]

Prove that the intersection of any collection of subspaces of \(V\) is a subspace of \(V\)

Is the operation of addition on the subspaces of \(V\) commutative? Associative? (In other words, if \(U_{1}, U_{2}, U_{3}\) are subspaces of \(V\), is \\[ \left.U_{1}+U_{2}=U_{2}+U_{1} ? \text { Is }\left(U_{1}+U_{2}\right)+U_{3}=U_{1}+\left(U_{2}+U_{3}\right) ?\right) \\]

Give an example of a nonempty subset \(U\) of \(\mathbf{R}^{2}\) such that \(U\) is closed under scalar multiplication, but \(U\) is not a subspace of \(\mathbf{R}^{2}.\)

For each of the following subsets of \(\mathbf{F}^{3}\), determine whether it is a subspace of \(\mathbf{F}^{3}:\) (a) \(\quad\left\\{\left(x_{1}, x_{2}, x_{3}\right) \in \mathbf{F}^{3}: x_{1}+2 x_{2}+3 x_{3}=0\right\\}\) (b) \(\quad\left\\{\left(x_{1}, x_{2}, x_{3}\right) \in \mathbf{F}^{3}: x_{1}+2 x_{2}+3 x_{3}=4\right\\}\) (c) \(\quad\left\\{\left(x_{1}, x_{2}, x_{3}\right) \in \mathbf{F}^{3}: x_{1} x_{2} x_{3}=0\right\\}\) (d) \(\quad\left\\{\left(x_{1}, x_{2}, x_{3}\right) \in \mathbf{F}^{3}: x_{1}=5 x_{3}\right\\}\)
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