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Chapter 6: Problem 81

Solve each equation in the complex number system. Express solutions in polar and rectangular form. $$x^{6}-1=0$$

Short Answer

Expert verified
The solutions to \(x^{6}-1=0\) in rectangular form are \( x = 1, -1/2 + \sqrt{3}/2 i, -1/2 - \sqrt{3}/2 i, -1, 1/2 - \sqrt{3}/2 i, 1/2 + \sqrt{3}/2 i \), and in polar form are \( x = cos(0) + i sin(0), cos(\pi/3) + i sin(\pi/3), cos(2\pi/3) + i sin(2\pi/3), cos(\pi) + i sin(\pi), cos(4\pi/3) + i sin(4\pi/3), cos(5\pi/3) + i sin(5\pi/3) \).

Step by step solution

01

Understanding the equation

This equation can be factored as a difference of two squares, that is, \(x^{6}-1=0\) can be written as \((x^{3} - 1)(x^{3} + 1) = 0\). Now we can solve for both \(x^{3} - 1 = 0\) and \(x^{3} + 1 = 0\) to get the six solutions required.
02

Solve for \(x^{3} - 1 = 0\)

Adding 1 to both sides gives: \(x^{3} = 1\). Applying De Moivre's theorem, that is \(z^{n}=r^{n}(\cos(n\Theta)+i \sin(n\Theta))\), where \(r=1\) and \( \Theta=0 \), we get: \(x=1(\cos((2k\pi)/3)+i \sin((2k\pi)/3))\), where \(k\) can take values of 0, 1, and 2. Solving for each value of \(k\) gives us the three solutions in polar form.
03

Solve for \(x^{3} + 1 = 0\)

Subtracting 1 from both sides gives: \(x^{3} = -1\). Again, using De Moivre's theorem with \(r=1\) and \( \Theta=\pi \), we get: \(x=1(\cos(((2k\pi)+\pi)/3)+i \sin(((2k\pi)+\pi)/3))\), where \(k\) can take values of 0, 1, and 2. Solving for each value of \(k\) for this part gives the remaining three solutions in the polar form.
04

Convert to Rectangular form

Using Euler's formula, each these solutions in polar form can be converted into rectangular form, which is of the form \(x = a + bi \), where \(a\) is the real part and \(b\) is the imaginary part.

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