Chapter 14.5, Problem 1OE

To determine

To find: the value of ${\text{z}}^{\text{2}}$ and ${\text{z}}^5$ in polar form of equation.

Answer to Problem 1OE

  ${\text{z}}^2\text{=}4\left(\text{cos}8{\text{0}}^{\text{0}}\text{+isin}8{\text{0}}^{\text{0}}\right)$ and ${\text{z}}^{\text{5}}\text{=32}\left({\text{cos200}}^{\text{0}}{\text{+isin200}}^{\text{0}}\right)$ .

Explanation of Solution

Given:

  $\text{z=2}\left({\text{cos40}}^{\text{0}}{\text{+isin40}}^{\text{0}}\right)$ .

Concept used:

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

  ${\text{x}}^{\text{n}}\text{=}\left(\text{cos2knπ+isin2knπ}\right)$ .

Calculation:

According to the given:

  $\text{z=2}\left({\text{cos40}}^{\text{0}}{\text{+isin40}}^{\text{0}}\right)$ .

  ${\text{z}}^2\text{=}{\left(\text{2}\left({\text{cos40}}^{\text{0}}{\text{+isin40}}^{\text{0}}\right)\right)}^2$ .

  ${\text{z}}^{\text{5}}\text{=}{\left(\text{2}\left({\text{cos40}}^{\text{0}}{\text{+isin40}}^{\text{0}}\right)\right)}^{\text{5}}$ .

According to the $\text{Demoivre}$ theorem:

  ${\text{z}}^{\text{n}}{\text{=r}}^{\text{n}}\left(\text{cosnθ+isinnθ}\right)$ .

  ${\text{z}}^2{\text{=2}}^2\left(\text{cos2}\times {\text{40}}^{\text{0}}\text{+isin2}\times {\text{40}}^{\text{0}}\right)$ .

  ${\text{z}}^2\text{=}4\left(\text{cos}8{\text{0}}^{\text{0}}\text{+isin}8{\text{0}}^{\text{0}}\right)$ .

Similarly,

  ${\text{z}}^{\text{5}}\text{=}{\left(\text{2}\left({\text{cos40}}^{\text{0}}{\text{+isin40}}^{\text{0}}\right)\right)}^{\text{5}}$ .

  ${\text{z}}^{\text{5}}\text{=32}\left({\text{cos5×40}}^{\text{0}}{\text{+isin5×40}}^{\text{0}}\right)$ .

  ${\text{z}}^{\text{5}}\text{=32}\left({\text{cos200}}^{\text{0}}{\text{+isin200}}^{\text{0}}\right)$ .

Hence, value of ${\text{z}}^2\text{=}4\left(\text{cos}8{\text{0}}^{\text{0}}\text{+isin}8{\text{0}}^{\text{0}}\right)$ and ${\text{z}}^{\text{5}}\text{=32}\left({\text{cos200}}^{\text{0}}{\text{+isin200}}^{\text{0}}\right)$ .