Chapter 14.5, Problem 18WE

To determine

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

Answer to Problem 18WE

  $\text{1}\left({\text{cos30}}^{\text{0}}{\text{+isin30}}^{\text{0}}\right)\text{=}\frac{\sqrt{\text{3}}}{\text{2}}\text{+i}\frac{\text{1}}{\text{2}}$ .

  $\text{1}\left({\text{cos150}}^{\text{0}}{\text{+isin150}}^{\text{0}}\right)\text{=-}\frac{\sqrt{\text{3}}}{\text{2}}\text{+i}\frac{\text{1}}{\text{2}}$ .

  $\text{1}\left({\text{cos270}}^{\text{0}}{\text{+isin270}}^{\text{0}}\right)\text{=0-i}$ .

Explanation of Solution

Given:

  $\text{i}$ .

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 cube roots of $\text{i}$ is:

Equation in polar form:

Here from the given equation $\text{x=0}$ and $\text{y=i}$ .

  $\text{z=x+iy=rcosθ+irsinθ}$ .

  ${\text{r}}^{\text{2}}\text{=}\sqrt{{\text{x}}^{\text{2}}{\text{+y}}^{\text{2}}}$ .

  $\text{r=}\sqrt{0{\text{+1}}^{\text{2}}}$ .

  $\text{r=}1$ .

  $\text{x=rcosθ}$ and $\text{y=rsinθ}$ .

  $\text{cosθ=}\frac{\text{x}}{\text{r}}$ and $\text{sinθ=}\frac{\text{y}}{\text{r}}$ .

  $\text{cosθ=}\frac{\text{x}}{\text{r}}=0$ .

  $\text{sinθ=}\frac{\text{y}}{\text{r}}=\frac{1}{1}=1$ .

From unit circle the angle can be calculated as:

  ${\text{θ=90}}^0$ .

  ${\text{z}}^3{\text{=r}}^3\left(\text{cos3θ+isin3θ}\right)=1\left({\text{cos90}}^0{\text{+isin90}}^0\right)$

  $\text{r=1}{\text{and cos3θ=90}}^{\text{0}}{\text{,sin3θ=90}}^{\text{0}}$ .

  ${\text{3θ=90}}^{\text{0}}\text{+k}{\text{.360}}^{\text{0}}$ .

  ${\text{θ=30}}^{\text{0}}\text{+k}{\text{.120}}^{\text{0}}$

For $\text{k=0,1,2}$

  ${\text{θ=30}}^{\text{0}}{\text{,150}}^{\text{0}}\text{and}27{\text{0}}^{\text{0}}$ .

Hence three cube roots are:

  $\text{1}\left({\text{cos30}}^{\text{0}}{\text{+isin30}}^{\text{0}}\right)\text{=}\frac{\sqrt{\text{3}}}{\text{2}}\text{+i}\frac{\text{1}}{\text{2}}$ .

  $\text{1}\left({\text{cos150}}^{\text{0}}{\text{+isin150}}^{\text{0}}\right)\text{=-}\frac{\sqrt{\text{3}}}{\text{2}}\text{+i}\frac{\text{1}}{\text{2}}$ .

  $\text{1}\left({\text{cos270}}^{\text{0}}{\text{+isin270}}^{\text{0}}\right)\text{=0-i}$ .