Chapter 6.6, Problem 101E

To determine

To verify:

The following sentence:

The sixth root of $1$ is $-\frac{1}{2}\left(1+\sqrt{3}i\right)$ .

Answer to Problem 101E

Yes, $-\frac{1}{2}\left(1+\sqrt{3}i\right)$ is sixth root of $1$

Explanation of Solution

Given information:

  $-\frac{1}{2}\left(1+\sqrt{3}i\right)$

Formula used:

The DeMoivre’s theorem for complex numbers is given as:

If $z=r\left(\cos +i\sin \theta \right)$ is complex number it has positive integer $n$ then

  $\begin{array}{l}{z}^n={\left[r\left(\cos \theta +i\sin \theta \right)\right]}^n\\ z^n=r^n\left(\cos n\theta +i\sin n\theta \right)\end{array}$

Calculation:

The complex number in the trigonometric form is derived as:

  $-\frac{1}{2}\left(1+\sqrt{3}i\right)$

Modulus and argument of the complex number is calculated as:

  $\begin{array}{l}r=\sqrt{a^2+b^2}=-\frac{1}{2}\sqrt{1^2+{\left(\sqrt{3}\right)}^2}=-\frac{1}{2}\times 2=-1\\ \theta ={\tan }^{-1}\left(\frac{b}{a}\right)={\tan }^{-1}\left(\frac{\sqrt{3}}{1}\right)=\frac{\pi }{3}\end{array}$

By using DeMoivre’s theorem:

  $\begin{array}{l}{z}^n=r^n\left(\cos n\theta +i\sin n\theta \right)\\ {\left[-\frac{1}{2}\left(1+\sqrt{3}i\right)\right]}^6={\left(\left(-1\right)\left[\cos \left(\frac{\pi }{3}\right)+i\sin \left(\frac{\pi }{3}\right)\right]\right)}^6\\ {\left[-\frac{1}{2}\left(1+\sqrt{3}i\right)\right]}^6={\left(-1\right)}^6\left[\cos \left(\frac{\pi }{3}\times 6\right)+i\sin \left(\frac{\pi }{3}\times 6\right)\right]\\ {\left[-\frac{1}{2}\left(1+\sqrt{3}i\right)\right]}^6=\left[\cos \left(2\pi \right)+i\sin \left(2\pi \right)\right]\\ {\left[-\frac{1}{2}\left(1+\sqrt{3}i\right)\right]}^6=\left(1+0i\right)\\ {\left[-\frac{1}{2}\left(1+\sqrt{3}i\right)\right]}^6=1\end{array}$

Hence, the sixth root of $1$ is $-\frac{1}{2}\left(1+\sqrt{3}i\right)$ .