Chapter 8.3, Problem 86E

To determine

To calculate: To find the indicated roots and graph the roots in the complex plane

Answer to Problem 86E

The cube roots are ${\omega }_0=2^{\frac{1}{6}}\left[\cos \frac{\pi }{12}+i\sin \frac{\pi }{12}\right]$ , ${\omega }_1=2^{\frac{1}{6}}\left[\cos \frac{3\pi }{4}+i\sin \frac{3\pi }{4}\right]$ and ${\omega }_2=2^{\frac{1}{6}}\left[\cos \frac{17\pi }{12}+i\sin \frac{17\pi }{12}\right]$

Explanation of Solution

Given information: The cube roots of $1+i$

Formula Used:

Complex number is a number that can be expressed in the form of $a+bi$ , where $a$ and $b$ are real numbers

  $z=a+bi$

If $z=r\left(\cos \theta +isin\theta \right)$ and $n$ is a positive integer, then $z$ has the $n$ distinct $n^{th}$ roots

  ${\omega }_k=r^{\frac{1}{n}}\left[\cos \left(\frac{\theta +2\pi k}{n}\right)+i\sin \left(\frac{\theta +2\pi k}{n}\right)\right]$ for $k=0,1,2,\mathrm{3....},n-1$

Calculation:

Complex number is given as

  $z=1+i$

Polar form of complex number is

  $z=r\left(\cos \theta +isin\theta \right)--(1)$

Calculating the value of $r$ :

  $\begin{array}{l}r=\sqrt{{\left(1\right)}^2+{\left(1\right)}^2}\\ r=\sqrt{1+1}\\ r=\sqrt{2}\end{array}$

Calculating the value of ${\theta }_1$

  $\begin{array}{l}\tan \theta =\frac{1}{1}\\ \tan \theta =1\\ \theta ={\tan }^{-1}1\\ \theta =\frac{\pi }{4}\end{array}$

Hence, complex number in polar form is

  $z=\sqrt{2}\left(\cos \frac{\pi }{4}+isin\frac{\pi }{4}\right)$

In order to find cube roots of above complex number, apply nth Rule of Complex numbers

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

Here, $n=3$ and $z=\sqrt{2}\left(\cos \frac{\pi }{4}+isin\frac{\pi }{4}\right)$

Thus,

  $\begin{array}{l}{\omega }_k={\sqrt{2}}^{\frac{1}{3}}\left[\cos \left(\frac{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$4$}\right.+2\pi k}{3}\right)+i\sin \left(\frac{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$4$}\right.+2\pi k}{3}\right)\right]\\ {\omega }_k=2^{\frac{1}{6}}\left[\cos \left(\frac{\pi }{12}+\frac{2\pi k}{3}\right)+i\sin \left(\frac{\pi }{12}+\frac{2\pi k}{3}\right)\right]\end{array}$

Now, the value of $k$ is from $0$ to $n-1$

Thus,

  $\begin{array}{l}0\le k\le 3-1\\ 0\le k\le 2\end{array}$

Hence, $k=0,1,2$

When $k=0$

  $\begin{array}{l}{\omega }_0=2^{\frac{1}{6}}\left[\cos \left(\frac{\pi }{12}+\frac{2\pi \left(0\right)}{3}\right)+i\sin \left(\frac{\pi }{12}+\frac{2\pi \left(0\right)}{3}\right)\right]\\ {\omega }_0=2^{\frac{1}{6}}\left[\cos \left(\frac{\pi }{12}+0\right)+i\sin \left(\frac{\pi }{12}+0\right)\right]\\ {\omega }_0=2^{\frac{1}{6}}\left[\cos \frac{\pi }{12}+i\sin \frac{\pi }{12}\right]\end{array}$

When $k=1$

  $\begin{array}{l}{\omega }_1=2^{\frac{1}{6}}\left[\cos \left(\frac{\pi }{12}+\frac{2\pi \left(1\right)}{3}\right)+i\sin \left(\frac{\pi }{12}+\frac{2\pi \left(1\right)}{3}\right)\right]\\ {\omega }_1=2^{\frac{1}{6}}\left[\cos \left(\frac{\pi }{12}+\frac{2\pi }{3}\right)+i\sin \left(\frac{\pi }{12}+\frac{2\pi }{3}\right)\right]\\ {\omega }_1=2^{\frac{1}{6}}\left[\cos \frac{9\pi }{12}+i\sin \frac{9\pi }{12}\right]\\ {\omega }_1=2^{\frac{1}{6}}\left[\cos \frac{3\pi }{4}+i\sin \frac{3\pi }{4}\right]\end{array}$

When $k=2$

  $\begin{array}{l}{\omega }_2=2^{\frac{1}{6}}\left[\cos \left(\frac{\pi }{12}+\frac{2\pi \left(2\right)}{3}\right)+i\sin \left(\frac{\pi }{12}+\frac{2\pi \left(2\right)}{3}\right)\right]\\ {\omega }_2=2^{\frac{1}{6}}\left[\cos \left(\frac{\pi }{12}+\frac{4\pi }{3}\right)+i\sin \left(\frac{\pi }{12}+\frac{4\pi }{3}\right)\right]\\ {\omega }_2=2^{\frac{1}{6}}\left[\cos \frac{17\pi }{12}+i\sin \frac{17\pi }{12}\right]\end{array}$

Now, plotting the above points on the complex plane. All points lie on a circle of radius $r=2^{\frac{1}{6}}$

  

Conclusion:

Hence, cube roots are ${\omega }_0=2^{\frac{1}{6}}\left[\cos \frac{\pi }{12}+i\sin \frac{\pi }{12}\right]$ , ${\omega }_1=2^{\frac{1}{6}}\left[\cos \frac{3\pi }{4}+i\sin \frac{3\pi }{4}\right]$ and ${\omega }_2=2^{\frac{1}{6}}\left[\cos \frac{17\pi }{12}+i\sin \frac{17\pi }{12}\right]$