Chapter 6, Problem 142CR

To determine

To explain:

The product and division form of following figure:

  

Answer to Problem 142CR

The product form is $z_1{z}_2=-4$ and division form is $\frac{z_1}{z_2}=-\cos 2\theta +i\sin 2\theta$ .

Explanation of Solution

The complex numbers can be produced from the plot; the values of expression can be calculated.

  $\begin{array}{l}{z}_1=2\left(\cos \theta +i\sin \theta \right)\\ z_2=2\left(\cos \left(\pi -\theta \right)+i\sin \left(\pi -\theta \right)\right)\\ z_2=2\left(-\cos \theta +i\sin \theta \right)\end{array}$

The two complex numbers product can be determined as:

  $\begin{array}{l}{z}_1{z}_2=2\left(\cos \theta +i\sin \theta \right)\times 2\left(-\cos \theta +i\sin \theta \right)\\ z_1{z}_2=4\left(-{\cos }^2\theta +i\cos \theta \sin \theta -i\sin \theta \cos \theta +i^2{\sin }^2\theta \right)\\ z_1{z}_2=4\left(-{\cos }^2\theta -{\sin }^2\theta \right)\\ z_1{z}_2=-4\left(-{\cos }^2\theta +i{\sin }^2\theta \right)\\ z_1{z}_2=-4\end{array}$

The product form of given equation is $z_1{z}_2=-4$

The two complex numbers can be determined as:

  $\begin{array}{l}\frac{z_1}{z_2}=\frac{2\left(\cos \theta +i\sin \theta \right)}{2\left(-\cos \theta +i\sin \theta \right)}\\ \frac{z_1}{z_2}=\frac{\left(\cos \theta +i\sin \theta \right)\left(-\cos \theta -i\sin \theta \right)}{\left(-\cos \theta +i\sin \theta \right)\left(-\cos \theta -i\sin \theta \right)}\\ \frac{z_1}{z_2}=\frac{-\left({\cos }^2\theta +i^2{\sin }^2\theta +2i\sin \theta \cos \theta \right)}{\left({\cos }^2\theta +i\sin \theta \cos \theta -i\sin \theta \cos \theta -i^2{\sin }^2\theta \right)}\\ \frac{z_1}{z_2}=\frac{-\left({\cos }^2\theta -{\sin }^2\theta +2i\sin \theta \cos \theta \right)}{\left({\cos }^2\theta +{\sin }^2\theta \right)}\\ \frac{z_1}{z_2}=\frac{-\left(\cos 2\theta +i\sin 2\theta \right)}{1}\\ \frac{z_1}{z_2}=-\cos 2\theta +i\sin 2\theta \end{array}$

A division form of given equation is $\frac{z_1}{z_2}=-\cos 2\theta +i\sin 2\theta$ .