Chapter 14.5, Problem 8WE

To determine

To find: the power of the expression using de $\text{Moivre}$ theorem and write the solution in the form of $\text{x+yi}$ .

Answer to Problem 8WE

  ${\left(\text{-}\sqrt{3}\text{+i}\right)}^9\text{=-256-256}\sqrt{\text{3}}\text{i}$ .

Explanation of Solution

Given:

  ${\left(\text{-}\sqrt{3}\text{+i}\right)}^9$ .

Concept used:

Trigonometric form of complex number:

  $\text{z=r}\left(\text{cos}\varphi \text{+isin}\varphi \right)$............ $\left(\text{i}\right)$

Where $\text{r=}\sqrt{{\text{a}}^{\text{2}}{\text{+b}}^{\text{2}}}$ is modulus of the complex number.

De $\text{Moivre}$ theorem:

  ${\text{z}}^{\text{n}}{\text{=r}}^{\text{n}}\left(\text{cosn}\varphi \text{+isinn}\varphi \right)$ .

Calculation:

  ${\left(\text{-}\sqrt{3}\text{+i}\right)}^9$ .

It can be displayed in trigonometric form:

  $\text{z=r}\left(\text{cos}\varphi \text{+isin}\varphi \right)$............ $\left(\text{i}\right)$

Where $\text{r=}\sqrt{{\text{a}}^{\text{2}}{\text{+b}}^{\text{2}}}$ is modulus of the complex number.

  $\text{r=}\sqrt{{\left(\text{-}\sqrt{3}\right)}^{\text{2}}\text{+}{\left(1\right)}^{\text{2}}}\text{=2}$ .

Since, $\text{x=rcosθ}$ and $\text{y=rsinθ}$

  $\text{cosθ=}\frac{\text{-}\sqrt{\text{3}}}{\text{2}}$ . And $\text{sinθ=}\frac{\text{1}}{\text{2}}$

Therefore ${\text{θ=150}}^{\text{0}}$ .

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

Applying De $\text{Moivre}$ theorem:

  ${\text{z}}^{\text{n}}{\text{=r}}^{\text{n}}\left(\text{cosn}\varphi \text{+isinn}\varphi \right)$ .

  ${\left(\text{-}\sqrt{3}\text{+i}\right)}^9{\text{=2}}^9{\left(\text{cos}\left({\text{150}}^{\text{0}}\right)\text{+isin}\left({\text{150}}^{\text{0}}\right)\right)}^9$ .

  $\Rightarrow 512\left(\text{cos}\left(9\times 15{\text{0}}^{\text{0}}\right)\text{+isin}\left(9\times 15{\text{0}}^{\text{0}}\right)\right)$ .

  $\Rightarrow 512\left(\text{cos}\left({1350}^{\text{0}}\right)\text{+isin}\left({1350}^{\text{0}}\right)\right)$ .

  $\Rightarrow 512\left(-\frac{1}{2}\text{-}\frac{\sqrt{3}}{2}\right)$ .

  $\Rightarrow \text{512}\left(\text{-}\frac{\text{1}}{\text{2}}\text{-}\frac{\sqrt{\text{3}}}{\text{2}}\text{i}\right)$ .

  $\Rightarrow \text{-256-256}\sqrt{\text{3}}\text{i}$ .

Hence, ${\left(\text{-}\sqrt{3}\text{+i}\right)}^9\text{=-256-256}\sqrt{\text{3}}\text{i}$ .