Chapter 6, Problem 24CT

To determine

The indicated power of the complex numberin standard form using DeMoivre’s Theorem.

Answer to Problem 24CT

Theindicated power of the complex numberin standard form is $z=5832i$

Explanation of Solution

Given info:

  $z={\left[3-3i\right]}^6$

Formula used:

The trigonometric form of the complex number $z=a+ib$ is

  $z=r\left(\cos \theta +i\sin \theta \right)$

Where $r=\sqrt{a^2+b^2}$ and $\theta ={\tan }^{-1}\left(\frac{b}{a}\right)$

DeMoivre’s Theorem:

If $z=r\left(\cos \theta +i\sin \theta \right)$ is a complex number and n is a positive integer, then

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

Calculation:

We have

  $\begin{array}{l}a=3,b=-3\\ r=\sqrt{a^2+b^2}=\sqrt{3^2+{\left(-3\right)}^2}=3\sqrt{2}\\ \theta ={\tan }^{-1}\left(\frac{-3}{3}\right)=\frac{7\pi }{4}\\ z=r\left(\cos \theta +i\sin \theta \right)\\ 3-3i=3\sqrt{2}\left(\cos \frac{7\pi }{4}+i\sin \frac{7\pi }{4}\right)\\ z={\left[3-3i\right]}^6\\ z={\left[3\sqrt{2}\left(\cos \frac{7\pi }{4}+i\sin \frac{7\pi }{4}\right)\right]}^6\\ {\left(r\left(\cos \theta +i\sin \theta \right)\right)}^n=r^n\left(\cos n\theta +i\sin n\theta \right)\\ z={\left[3\sqrt{2}\left(\cos \frac{7\pi }{4}+i\sin \frac{7\pi }{4}\right)\right]}^6={\left(3\sqrt{2}\right)}^6\left(\cos \left(6\times \frac{7\pi }{4}\right)+i\sin \left(6\times \frac{7\pi }{4}\right)\right)\\ z=5832\left(\cos \frac{21\pi }{2}+i\sin \frac{21\pi }{2}\right)=5832\left(\cos \frac{\pi }{2}+i\sin \frac{\pi }{2}\right)\\ z=5832i\end{array}$

The indicated power of the complex numberin standard form is $z=5832i$

Conclusion:

Thus,the indicated power of the complex numberin standard form is $z=5832i$