Chapter 6.5, Problem 64E

To determine

The powers $z,z^2,z^3\text{and }{z}^4$ graphically. Describe the pattern.

Answer to Problem 64E

The powers $z,z^2,z^3\text{and }{z}^4$ are found out and represented graphically, the absolute value of each is 1, and consecutivepowers of z are each ${60}^0$ apart.

Explanation of Solution

Given info:

  $z=\frac{1}{2}\left(1+\sqrt{3}i\right)$

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

  $\begin{array}{l}z=\frac{1}{2}\left(1+\sqrt{3}i\right)=\frac{1}{2}+\frac{\sqrt{3}}{2}i\\ a=\frac{1}{2},b=\frac{\sqrt{3}}{2}\\ r=\sqrt{a^2+b^2}=\sqrt{{\left(\frac{1}{2}\right)}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2}=1\\ \theta ={\tan }^{-1}\left(\frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}\right)={60}^0\\ z=r\left(\cos \theta +i\sin \theta \right)\\ \frac{1}{2}+\frac{\sqrt{3}}{2}i=\cos {60}^0+i\sin {60}^0\\ z=\cos {60}^0+i\sin {60}^0\\ z^2=\cos \left(2\times {60}^0\right)+i\sin \left(2\times {60}^0\right)=\cos {120}^0+i\sin {120}^0=-\frac{1}{2}+\frac{\sqrt{3}}{2}i\\ z^3=\cos \left(3\times {60}^0\right)+i\sin \left(3\times {60}^0\right)=\cos {180}^0+i\sin {180}^0=-1\\ z^4=\cos \left(4\times {60}^0\right)+i\sin \left(4\times {60}^0\right)=\cos {240}^0+i\sin {240}^0=-\frac{1}{2}-\frac{\sqrt{3}}{2}i\end{array}$

Graphical representation

  

The absolute value of each is 1, and consecutivepowers of z are each ${60}^0$ apart.

Conclusion:

Thus,thepowers $z,z^2,z^3\text{and }{z}^4$ are found out and represented graphically, the absolute value of each is 1, and consecutivepowers of z are each ${60}^0$ apart.