Chapter 14.5, Problem 16WE

To determine

To find: the required roots in $\text{x+iy}$ form.

Answer to Problem 16WE

  $\text{1+0i}$ , $\frac{\text{1}}{\sqrt{\text{2}}}\text{+i}\frac{\text{1}}{\sqrt{\text{2}}}$ , $\text{0+i}$ , $\text{-}\frac{\sqrt{\text{2}}}{\text{2}}\text{+i}\frac{\sqrt{\text{2}}}{\text{2}}$ , $\text{-1+0i}$ , $\text{-}\frac{\sqrt{\text{2}}}{\text{2}}\text{-i}\frac{\sqrt{\text{2}}}{\text{2}}$ , $\text{0-i}$ and $\frac{\sqrt{\text{2}}}{\text{2}}\text{-i}\frac{\sqrt{\text{2}}}{\text{2}}$ .

Explanation of Solution

Given:

The eighth roots of unity.

Concept used:

If $\text{x}$ is the an $\text{nth}$ root of unity, then it satisfies the relation:

  ${\text{x}}^{\text{n}}\text{=1}$ .

  $\text{=cos0+isin0}$ .

  ${\text{x}}^{\text{n}}\text{=cos2kπ+isin2kπ}$ where $\text{k}$ is an integer.

Taking $\text{nth}$ root on both sides:

  ${\left({\text{x}}^{\text{n}}\right)}^{\frac{\text{1}}{\text{n}}}\text{=}{\left(\text{cos2kπ+isin2kπ}\right)}^{\frac{\text{1}}{\text{n}}}$ .

  $\text{x=}{\left(\text{cos2kπ+isin2kπ}\right)}^{\frac{\text{1}}{\text{n}}}$ .

According $\text{de moivre}$ s law $\text{x=}{\left(\text{cos2kπ+isin2kπ}\right)}^{\frac{\text{1}}{\text{n}}}$ can be written as:

In radian: $\text{x=}\left(\text{cos}\frac{\text{2kπ}}{\text{n}}\text{+isin}\frac{\text{2kπ}}{\text{n}}\right)$ .

In degree: $\text{x=}\left(\text{cos}\frac{\text{360k}}{\text{n}}\text{+isin}\frac{\text{360k}}{\text{n}}\right)$ .

Calculation:

The eighthroot of unity.

Using this formula:

  $\text{x=}\left(\text{cos}\frac{\text{360k}}{\text{n}}\text{+isin}\frac{\text{360k}}{\text{n}}\right)$ .

When $\text{k=0}$ ;

  $\text{x=}\left(\text{cos}\frac{\text{360×0}}{8}\text{+isin}\frac{\text{360×0}}{8}\right)\text{=1+0i}$ .

When $\text{k=1}$ :

  $\text{x=}\left(\text{cos}\frac{\text{360×1}}{\text{8}}\text{+isin}\frac{\text{360×1}}{\text{8}}\right){\text{=cos45}}^{\text{0}}{\text{+isin45}}^{\text{0}}\text{=}\frac{\text{1}}{\sqrt{\text{2}}}\text{+i}\frac{\text{1}}{\sqrt{\text{2}}}$ .

When $\text{k=2}$ :

  $\text{x=}\left(\text{cos}\frac{\text{360×2}}{\text{8}}\text{+isin}\frac{\text{360×2}}{\text{8}}\right){\text{=cos90}}^{\text{0}}{\text{+isin90}}^{\text{0}}\text{=0+i}$ .

When $\text{k=3}$

  $\text{x=}\left(\text{cos}\frac{\text{360×3}}{\text{8}}\text{+isin}\frac{\text{360×3}}{\text{8}}\right){\text{=cos135}}^{\text{0}}{\text{+isin135}}^{\text{0}}\text{=-}\frac{\sqrt{\text{2}}}{\text{2}}\text{+i}\frac{\sqrt{\text{2}}}{\text{2}}$ .

When $\text{k=4}$

  $\text{x=}\left(\text{cos}\frac{\text{360×4}}{\text{8}}\text{+isin}\frac{\text{360×4}}{\text{8}}\right){\text{=cos180}}^{\text{0}}{\text{+isin180}}^{\text{0}}\text{=-1+0i}$ .

When $\text{k=5}$

  $\text{x=}\left(\text{cos}\frac{\text{360×5}}{\text{8}}\text{+isin}\frac{\text{360×5}}{\text{8}}\right){\text{=cos225}}^{\text{0}}{\text{+isin225}}^{\text{0}}\text{=-}\frac{\sqrt{\text{2}}}{\text{2}}\text{-i}\frac{\sqrt{\text{2}}}{\text{2}}$ .

When $\text{k=6}$

  $\text{x=}\left(\text{cos}\frac{\text{360×6}}{\text{8}}\text{+isin}\frac{\text{360×6}}{\text{8}}\right){\text{=cos270}}^{\text{0}}{\text{+isin270}}^{\text{0}}\text{=0-i}$ .

When $\text{k=7}$

  $\text{x=}\left(\text{cos}\frac{\text{360×7}}{\text{8}}\text{+isin}\frac{\text{360×7}}{\text{8}}\right){\text{=cos315}}^{\text{0}}{\text{+isin315}}^{\text{0}}\text{=}\frac{\sqrt{\text{2}}}{\text{2}}\text{-i}\frac{\sqrt{\text{2}}}{\text{2}}$ .

Hence, the eighthroot of unity in $\text{x+iy}$ forms are:

  $\text{1+0i}$ , $\frac{\text{1}}{\sqrt{\text{2}}}\text{+i}\frac{\text{1}}{\sqrt{\text{2}}}$ , $\text{0+i}$ , $\text{-}\frac{\sqrt{\text{2}}}{\text{2}}\text{+i}\frac{\sqrt{\text{2}}}{\text{2}}$ , $\text{-1+0i}$ , $\text{-}\frac{\sqrt{\text{2}}}{\text{2}}\text{-i}\frac{\sqrt{\text{2}}}{\text{2}}$ , $\text{0-i}$ and $\frac{\sqrt{\text{2}}}{\text{2}}\text{-i}\frac{\sqrt{\text{2}}}{\text{2}}$ .