Chapter 9, Problem 62SGA

To determine

To Find: The power and obtain the rectangular form.

Answer to Problem 62SGA

The required rectangular form is $-16-16i$ .

Explanation of Solution

Given:

The given expression is ${\left(-2-2i\right)}^3$ .

Calculation:

Consider the complex number is ${\left(-2-2i\right)}^3$ .

Then, the polar form is obtained as, first calculate the amplitude.

  $\begin{array}{c}r=\sqrt{a^2+b^2}\\ =\sqrt{{\left(-2\right)}^2+{\left(-2\right)}^2}\\ =\sqrt{4+4}\\ =\sqrt{8}\end{array}$

The phase is obtained as,

  $\begin{array}{c}\theta =\mathrm{Arctan}\left(\frac{b}{a}\right)\\ =\arctan \left(\frac{-2}{-2}\right)+\pi \\ =\frac{5\pi }{4}\end{array}$

Thus, the polar form is $-2-2i=\sqrt{8}\left[\cos \frac{5\pi }{4}+i\sin \frac{\frac{5\pi }{4}\pi }{4}\right]$ .

The power of 7 by the De Moivre’s theorem is,

  $\begin{array}{c}{\left(-2-2i\right)}^3=\sqrt{8}{\left[\cos \frac{5\pi }{4}+i\sin \frac{5\pi }{4}\right]}^3\\ ={\left(\sqrt{8}\right)}^4\left(\cos 3\frac{5\pi }{4}+i\sin 3\frac{5\pi }{4}\right)\\ =16\sqrt{2}\left(\cos 3\frac{5\pi }{4}+i\sin 3\frac{5\pi }{4}\right)\\ =-16-16i\end{array}$