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Chapter 6: Problem 73

Find all the complex roots. Write roots in rectangular form. If necessary, round to the nearest tenth. The complex cube roots of 1

Short Answer

Expert verified
The cube roots of 1 in rectangular form are 1, \(-\frac{1}{2} + i\frac{\sqrt{3}}{2}\), and \(-\frac{1}{2} - i\frac{\sqrt{3}}{2}\).

Step by step solution

01

Understand the concept

The n complex roots of a number lie on a circle in the complex plane and are evenly distributed. The cube roots of 1 are 3 complex numbers that lie on a circle of radius 1 and are evenly distributed every \(\frac{2\pi}{3}\) radians.
02

Calculate roots

The cube roots of 1 can be calculated using the formula for the nth root of a complex number. In polar form, these roots are as follows: for k=0, 1, 2 - \(r=\sqrt[3]{1}=1\),\(\theta = \frac{2\pi k}{3}\).
03

Write results

Substituting k=0, 1, 2, the roots are \(re^{i\theta}=1, e^{\frac{2i\pi}{3}}, e^{\frac{4i\pi}{3}}\) respectively.
04

Convert roots to rectangular form

The rectangular form of a complex root \(re^{i\theta}\) is \(r(cos\theta + isin\theta)\). By replacing r, and 胃 for each root, rectangular form roots are: \(1 (cos0 + isin0) = 1, 1 (cos\frac{2\pi}{3} + isin\frac{2\pi}{3}) = -\frac{1}{2} + i\frac{\sqrt{3}}{2}, 1 (cos\frac{4\pi}{3} + isin\frac{4\pi}{3}) = -\frac{1}{2} - i\frac{\sqrt{3}}{2}\).

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