Chapter I, Problem 37E

To determine

To find: The roots of the eighth root of 1 and sketch the roots in the complex plane.

Answer to Problem 37E

The roots of eighth root of 1 are $w_k=1^{\frac{1}{n}}\left[\cos \left(\frac{0+2k\pi }{8}\right)+i\sin \left(\frac{0+2k\pi }{8}\right)\right]=\cos \frac{k\pi }{8}+i\sin \frac{k\pi }{8}$  where $k=0,1,2,\dots ,7$.

Explanation of Solution

Theorem used:

Roots of a complex number:

Let $z=r\left(\cos \theta +i\sin \theta \right)$ and n be a positive integer. Then $z$ has the $n$ distinct nth roots $w_k=r^{\frac{1}{n}}\left[\cos \left(\frac{\theta +2k\pi }{n}\right)+i\sin \left(\frac{\theta +2k\pi }{n}\right)\right]$ where $k=0,1,2,\dots ,n-1$.

Calculation:

Rewrite the complex number 1 in polar form.

The polar form of the complex number $z=a+bi$ is $z=r\left(\cos \theta +i\sin \theta \right)$ where $r=\left|z\right|=\sqrt{a^2+b^2}$  and $\tan \theta =\frac{b}{a}$.

Consider the complex number 1.

Obtain the argument of the complex number 1.

$\begin{array}{c}\tan \theta =\frac{0}{1}\\ =0\end{array}$

Thus, the argument of argument of the complex number 1 is $\theta ={\tan }^{-1}\left(0\right)=0$

Obtain the modulus of the complex number 1.

$\begin{array}{c}r=\left|1\right|\\ =\sqrt{1^2+{\left(0\right)}^2}\\ =\sqrt{1}\\ =1\end{array}$

Thus, the value of $r=1$.

Therefore, the polar form of the complex number 1 is $1=\left(\cos 0+i\sin 0\right)$.

By the above theorem, the roots of eighth root of 1 are $w_k=1^{\frac{1}{n}}\left[\cos \left(\frac{0+2k\pi }{8}\right)+i\sin \left(\frac{0+2k\pi }{8}\right)\right]=\cos \frac{k\pi }{8}+i\sin \frac{k\pi }{8}$ where $k=0,1,2,\dots ,7$.

Use online calculator to sketch the roots in the complex plane as shown below in Figure 1.

From figure 1, it is observed that all roots of eighth root of 1 form a circle on complex plane.