Chapter 6, Problem 40CLT

To determine

Thecube roots of $1$ using formula for roots.

Answer to Problem 40CLT

Thecube roots of $1$ are $z_0=10\left(\cos 0+i\sin 0\right),z_1=10\left(\cos \frac{2\pi }{3}+i\sin \frac{2\pi }{3}\right)\text{ and }{z}_2=10\left(\cos \frac{4\pi }{3}+i\sin \frac{4\pi }{3}\right)$

Explanation of Solution

Given info:

  $\sqrt[3]{1}$

Formula used:

The trigonometric form of the complex number $z=a+ib$ is

  $z=r\left(\cos \theta +i\sin \theta \right)$

Where $r=\sqrt{a^2+b^2}$ and $\theta ={\tan }^{-1}\left(\frac{b}{a}\right)$

For a positive integer n the complex number $z=r\left(\cos \theta +i\sin \theta \right)$ has exactly n distinct nth roots given by

  $z_k=\sqrt[n]{r}\left(\cos \left(\frac{\theta +2\pi k}{n}\right)+i\sin \left(\frac{\theta +2\pi k}{n}\right)\right)$

where k = 0, 1, 2, . . . ,,n -1.

Calculation:

We have

  $\begin{array}{l}a=1,b=0\\ r=\sqrt{a^2+b^2}=\sqrt{1^2+0^2}=1\\ \theta ={\tan }^{-1}\left(\frac{0}{1}\right)=0\\ 1=\cos 0+i\sin 0\\ z_k=\sqrt[n]{r}\left(\cos \left(\frac{\theta +2\pi k}{n}\right)+i\sin \left(\frac{\theta +2\pi k}{n}\right)\right)\\ n=3,r=1,\theta =0\\ z_0=\sqrt[3]{1}\left(\cos \left(\frac{0+2\pi \times 0}{3}\right)+i\sin \left(\frac{0+2\pi \times 0}{3}\right)\right)\\ z_0=\cos 0+i\sin 0\\ z_1=\sqrt[3]{1}\left(\cos \left(\frac{0+2\pi \times 1}{3}\right)+i\sin \left(\frac{0+2\pi \times 1}{3}\right)\right)\\ z_1=\cos \frac{2\pi }{3}+i\sin \frac{2\pi }{3}\\ z_2=\sqrt[3]{1}\left(\cos \left(\frac{0+2\pi \times 2}{3}\right)+i\sin \left(\frac{0+2\pi \times 2}{3}\right)\right)\\ z_2=\cos \frac{4\pi }{3}+i\sin \frac{4\pi }{3}\end{array}$

Thecube roots of $1$ are $z_0=\cos 0+i\sin 0,z_1=\cos \frac{2\pi }{3}+i\sin \frac{2\pi }{3}\text{ and }{z}_2=\cos \frac{4\pi }{3}+i\sin \frac{4\pi }{3}$

Conclusion:

Thus,the cube roots of $1$ are $z_0=\cos 0+i\sin 0,z_1=\cos \frac{2\pi }{3}+i\sin \frac{2\pi }{3}\text{ and }{z}_2=\cos \frac{4\pi }{3}+i\sin \frac{4\pi }{3}$