Chapter 6, Problem 27CT

To determine

To find: The power of the complex number and the result in the standard form.

Answer to Problem 27CT

The standard form of the given complex number is $5832i$ and the indicated power of the complex number is $6$ .

Explanation of Solution

Given:

The given complex number is ${\left(3-3i\right)}^6$ .

Calculation:

Consider the given function is,

  ${\left(3-3i\right)}^6$

The magnitude of the given complex number is,

  $\begin{array}{c}r=\sqrt{3^2+{\left(-3\right)}^2}\\ =\sqrt{18}\\ =3\sqrt{2}\end{array}$

The phase angle of the given complex number is,

  $\begin{array}{c}\theta =\arctan \left(\frac{-3}{3}\right)\\ =\frac{7\pi }{4}\end{array}$

Then, the trigonometric form of the given function is,

  ${\left(3-3i\right)}^6=\left[3\sqrt{2}\left(\cos \frac{7\pi }{4}+i\sin \frac{7\pi }{4}\right)\right]`$

Consider the Demoviers’s Theorem is,

  ${\left(\cos \theta +i\sin \theta \right)}^n=\cos n\theta +i\sin n\theta$

Then,

  $\begin{array}{c}{\left(3-3i\right)}^6=\left[3\sqrt{2}\left(\cos \frac{7\pi }{4}+i\sin \frac{7\pi }{4}\right)\right]\\ ={\left(3\sqrt{2}\right)}^6\left(\cos 6\left(\frac{7\pi }{4}\right)+i\sin 6\left(\frac{7\pi }{4}\right)\right)\\ =5832\left(\cos \frac{21\pi }{2}+i\sin \frac{21\pi }{2}\right)\\ =5832\left(0+i\left(1\right)\right)\end{array}$

Solve further as,

  ${\left(3-3i\right)}^6=5832i$