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Chapter 6: Problem 30

Explain why the six distinct complex numbers that are sixth roots of 1 are the vertices of a regular hexagon inscribed in the unit circle.

Short Answer

Expert verified
The six distinct complex numbers that are sixth roots of 1 can be written in polar form as \(z_k = \cos(\frac{2\pi k}{6}) + i\sin(\frac{2\pi k}{6})\) for \(k = 0, 1, \dots, 5\). These roots have a magnitude of 1 and angles evenly spaced at multiples of \(\frac{\pi}{3}\), thus forming a regular hexagon inscribed in the unit circle.

Step by step solution

01

Understanding sixth roots of 1

In this exercise, we are looking for complex numbers z such that z^6 = 1. We will first find these complex numbers in their polar form.
02

Find the polar form of 1

Note that 1 can be written in the polar form as \(r(\cos(\theta) + i\sin(\theta))\), where r is the magnitude and \(\theta\) is the angle. Since the magnitude of 1 is equal to 1, we have r = 1. The angle \(\theta\) can be 0 (the first root) and then we can find other roots by adding integer multiples of \(\frac{2\pi}{6}\) to 0.
03

Find the six complex roots of 1

Now, we will find the six complex roots by plugging in integer values for k in the following equation: \(z_k = \cos(\frac{2\pi k}{6}) + i\sin(\frac{2\pi k}{6}) \text{ for } k = 0, 1, \dots, 5\) Plugging in the values of k, we have: \(z_0 = \cos(0) + i\sin(0) = 1\) \(z_1 = \cos(\frac{\pi}{3}) + i\sin(\frac{\pi}{3}) = \frac{1}{2} + i\frac{\sqrt{3}}{2}\) \(z_2 = \cos(\frac{2\pi}{3}) + i\sin(\frac{2\pi}{3}) = -\frac{1}{2} + i\frac{\sqrt{3}}{2}\) \(z_3 = \cos(\pi) + i\sin(\pi) = -1\) \(z_4 = \cos(\frac{4\pi}{3}) + i\sin(\frac{4\pi}{3}) = -\frac{1}{2} - i\frac{\sqrt{3}}{2}\) \(z_5 = \cos(\frac{5\pi}{3}) + i\sin(\frac{5\pi}{3}) = \frac{1}{2} - i\frac{\sqrt{3}}{2}\)
04

Understand the geometry of the complex roots

In the complex plane, the polar representation describes points in terms of the magnitude (radius from the origin) and angle from the positive x-axis (real axis). The magnitude of the sixth roots is always 1 (since they lie on the unit circle), and their angles are evenly spaced at multiples of \(\frac{\pi}{3}\). This means that they form a regular hexagon inscribed in the unit circle. So, the six distinct complex numbers that are sixth roots of 1 are indeed vertices of a regular hexagon inscribed in the unit circle.

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Most popular questions from this chapter

Write each expression in the form \(a+b i,\) where \(a\) and \(b\) are real numbers. $$ (\sqrt{5}-\sqrt{7} i)^{2} $$

Write each expression in the form \(a+b i,\) where \(a\) and \(b\) are real numbers. $$ (1+3 i)-(6-5 i) $$

Show that multiplication of complex numbers is commutative, meaning that $$ w z=z w $$ for all complex numbers \(w\) and \(z\).

Write out a table showing the values of \(i^{n}\) with \(n\) ranging over the integers from 1 to 12 . Describe the pattern that emerges.

Show that multiplication of complex numbers is associative, meaning that $$ u(w z)=(u w) z $$ for all complex numbers \(u, w,\) and \(z\).
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