Chapter 14.5, Problem 2WE

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 2WE

  ${\left(\text{-1-i}\right)}^{\text{6}}\text{=0x-8i}$ .

Explanation of Solution

Given:

  ${\left(\text{-1-i}\right)}^{\text{6}}$ .

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:

  ${\left(\text{-1-i}\right)}^{\text{6}}$ .

It can be displayed in trigonometric form:

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

Where $\text{r=}\sqrt{{\left(\text{-1}\right)}^{\text{2}}\text{+}{\left(\text{-1}\right)}^{\text{2}}}\text{=}\sqrt{\text{2}}$ is modulus of the complex number.

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

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

Therefore ${\text{θ=225}}^{\text{0}}$ .

  $\left(\text{-1-i}\right)\text{=}\sqrt{\text{2}}\left({\text{cos225}}^{\text{0}}{\text{+isin225}}^{\text{0}}\right)$ .

Applying De $\text{Moivre}$ theorem:

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

  ${\left(\text{-1-i}\right)}^{\text{6}}\text{=}{\left(\sqrt{\text{2}}\right)}^{\text{6}}{\left(\text{cos}\left({\text{225}}^{\text{0}}\right)\text{+isin}\left({\text{225}}^{\text{0}}\right)\right)}^{\text{6}}$ .

  $\Rightarrow 8\left(\text{cos}\left(6\times {225}^{\text{0}}\right)\text{+isin}\left(\text{6}\times {\text{225}}^{\text{0}}\right)\right)$ .

  $\Rightarrow 8\left(\text{cos}\left({1350}^{\text{0}}\right)\text{+isin}\left({1350}^{\text{0}}\right)\right)$ .

  $\Rightarrow \text{8}\left(\text{0+i}\left(\text{-1}\right)\right)\text{=-8i}$ .

Hence, ${\left(\text{-1-i}\right)}^{\text{6}}\text{=0x-8i}$ .