Chapter 6.6, Problem 67E

To determine

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

Answer to Problem 67E

Theindicated power of the complex numberin standard form is $z=-597-122i$

Explanation of Solution

Given info:

  $z={\left[3-2i\right]}^5$

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

  $\begin{array}{l}a=3,b=-2\\ r=\sqrt{a^2+b^2}=\sqrt{3^2+{\left(-2\right)}^2}=\sqrt{13}\\ \theta ={\tan }^{-1}\left(\frac{-2}{3}\right)={326.31}^0\\ z=r\left(\cos \theta +i\sin \theta \right)\\ 3-2i=\sqrt{13}\left(\cos {326.31}^0+i\sin {326.31}^0\right)\\ z={\left[3-2i\right]}^5\\ z={\left[\sqrt{13}\left(\cos {326.31}^0+i\sin {326.31}^0\right)\right]}^5\\ {\left(r\left(\cos \theta +i\sin \theta \right)\right)}^n=r^n\left(\cos n\theta +i\sin n\theta \right)\\ z={\left[\sqrt{13}\left(\cos {326.31}^0+i\sin {326.31}^0\right)\right]}^5={\left(\sqrt{13}\right)}^5\left(\cos \left(5\times {326.31}^0\right)+i\sin \left(5\times {326.31}^0\right)\right)\\ z=169\sqrt{13}\left(\cos {1631.55}^0+i\sin {1631.55}^0\right)\\ z=-597-122i\end{array}$

The indicated power of the complex numberin standard form is $z=-597-122i$

Conclusion:

Thus,theindicated power of the complex numberin standard form is $z=-597-122i$