Chapter 8, Problem 41RE

To determine

The indicated roots. Find the cube roots of $1$

Answer to Problem 41RE

The square roots of $1$ is $1,-1,\frac{1+i\sqrt{3}}{2},\frac{-1+i\sqrt{3}}{2},\frac{-1-i\sqrt{3}}{2},\frac{1-i\sqrt{3}}{2}$ .

Explanation of Solution

Given:

  $1$

Concept Used:

Indicated roots, if

  $z=r\left(\cos \theta +i\sin \theta \right)$ is a complex number in polar form and $n$ is a positive integer then $z$ has $n$ distinct roots, $w_0,w_1,w\dots ,w_{n-1}$ as

  $w_k=r^{1/n}\left[\cos \left(\frac{\theta +2k\pi }{n}\right)+i\sin \left(\frac{\theta +2k\pi }{n}\right)\right]$

Where, $k=0,1,2,\dots ,n-1.$

Calculations:

The given complex number is $4+4\sqrt{3}i$ .

Now by comparing the real and imaginary parts is $a=1,b=0$

Squaring and adding real and imaginary parts, find the value of $r$

  $r=\sqrt{1^2+{\left(0\right)}^2}=\sqrt{1}=1$

Angle can be found by dividing imaginary by real part of the complex number, quadrant can be using sign of real $x$ and imaginary $y$ part, so the angle can be found accordingly.

  $\theta ={\tan }^{-1}\left(\frac{0}{1}\right)={\tan }^{-1}\left(0\right)=0$ as $x:+ve,y:+ve,\text{so}1st\text{quadrant}$

After conversion to the polar form the equation is,

  $1=1\left(\cos 0+i\sin 0\right)$

Here $n=6,k=0$ for the first root.

  $\begin{array}{l}{w}_1=1^{1/6}\left(\cos \left(\frac{0+0\times 2\pi }{6}\right)+i\sin \left(\frac{0+0\times 2\pi }{6}\right)\right)\\ w_1=1\left(\cos 0+i\sin 0\right)=1\end{array}$

  $k=1,$ for second root

  $\begin{array}{l}{w}_2=6^{1/6}\left(\cos \left(\frac{0+1\times 2\pi }{6}\right)+i\sin \left(\frac{0+1\times 2\pi }{6}\right)\right)\\ w_2=1\left(\cos \frac{\pi }{3}+i\sin \frac{\pi }{3}\right)=\frac{1+i\sqrt{3}}{2}\end{array}$

  $k=2,$ for third root

  $\begin{array}{l}{w}_3=1^{1/6}\left(\cos \left(\frac{0+2\times 2\pi }{6}\right)+i\sin \left(\frac{0+2\times 2\pi }{6}\right)\right)\\ w_3=1\left(\cos \frac{2\pi }{9}+i\sin \frac{2\pi }{9}\right)=\frac{-1+i\sqrt{3}}{2}\end{array}$

  $k=3,$ for fourth root

  $\begin{array}{l}{w}_4=1^{1/6}\left(\cos \left(\frac{0+3\times 2\pi }{6}\right)+i\sin \left(\frac{0+3\times 2\pi }{6}\right)\right)\\ w_4=1\left(\cos \pi +i\sin \pi \right)=-1\end{array}$

  $k=4,$ for fifth root

  $\begin{array}{l}{w}_5=1^{1/6}\left(\cos \left(\frac{0+4\times 2\pi }{6}\right)+i\sin \left(\frac{0+4\times 2\pi }{6}\right)\right)\\ w_5=1\left(\cos \frac{4\pi }{3}+i\sin \frac{4\pi }{3}\right)=\frac{-1-i\sqrt{3}}{2}\end{array}$

  $k=5,$ for sixth root

  $\begin{array}{l}{w}_6=1^{1/6}\left(\cos \left(\frac{0+5\times 2\pi }{6}\right)+i\sin \left(\frac{0+5\times 2\pi }{6}\right)\right)\\ w_6=1\left(\cos \frac{5\pi }{3}+i\sin \frac{5\pi }{3}\right)=\frac{1-i\sqrt{3}}{2}\end{array}$

Conclusion:

Hence, the square roots of $1$ is $1,-1,\frac{1+i\sqrt{3}}{2},\frac{-1+i\sqrt{3}}{2},\frac{-1-i\sqrt{3}}{2},\frac{1-i\sqrt{3}}{2}$ .