Chapter 6.6, Problem 100E

To determine

To graph:

The power $z,z^2,z^3,z^4$ in graphical from and to describe the pattern

Explanation of Solution

Given information:

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

Graph:

  

Interpretation:

In the first step consider the complex number,

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

Then find $z^2,z^3,z^4$ and graph them on the same axes.

Now to find $z$ ,

Consider,

  $\begin{array}{l}z=\frac{1}{2}\left(1+\sqrt{3i}\right)\\ =\frac{1}{2}+\frac{\sqrt{3}}{2}i\\ z=\cos \left(\frac{\pi }{3}\right)+i\sin \left(\frac{\pi }{3}\right)\end{array}$

Now find $z^2$ ,

  $\begin{array}{l}{z}^2=\left(\cos \left(\frac{\pi }{3}\right)\right)+i\sin {\left(\frac{\pi }{3}\right)}^{{}^2}\\ =\cos \left(\frac{2\pi }{3}\right)+i\sin \left(\frac{2\pi }{3}\right)\left[\because {\left(\cos \theta +i\sin \theta \right)}^{\circ }=\cos n\theta +i\sin n\theta \right]\\ \\ z^2=-\frac{1}{2}+\frac{\sqrt{3}}{2}i\left[\because \cos \left(\frac{2\pi }{3}\right)=-\frac{1}{2},\sin \left(\frac{2\pi }{3}\right)=\frac{\sqrt{3}}{2}\right]\end{array}$

Similarly find the $z^3$ ,

  $\begin{array}{l}{z}^3=\left(\cos \left(\frac{\pi }{3}\right)\right)+i\sin {\left(\frac{\pi }{3}\right)}^3\\ =\cos \left(\pi \right)+i\sin \left(\pi \right)\left[\because {\left(\cos \theta +i\sin \theta \right)}^{\circ }=\cos n\theta +i\sin n\theta \right]\\ \\ z^3=-1\left[\because \cos \left(\pi \right)=-1,\sin \left(\pi \right)\right]\end{array}$

Now also find the $z^4$ ,

  $\begin{array}{l}{z}^4=\left(\cos \left(\frac{\pi }{3}\right)\right)+i\sin {\left(\frac{\pi }{3}\right)}^4\\ =\cos \left(\frac{4\pi }{3}\right)+i\sin \left(\frac{4\pi }{3}\right)\left[\because {\left(\cos \theta +i\sin \theta \right)}^{\circ }=\cos n\theta +i\sin n\theta \right]\\ \\ z^4=-\frac{1}{2}-\frac{\sqrt{3}}{2}i\left[\because \cos \left(\frac{4\pi }{3}\right)=-1,\sin \left(\frac{4\pi }{3}\right)=0\right]\end{array}$

Now sketch the graph from the above obtained values:

  

From the figure, all the points are at unit distance and at an angle of ${60}^{\circ }$ from each other.