Chapter 6.6, Problem 43E

To determine

Quotient of the complex numbers in trigonometric form.

Answer to Problem 43E

Thequotient of the complex numbers in trigonometric form is $z=\cos \left(\frac{2\pi }{3}\right)+i\sin \left(\frac{2\pi }{3}\right)$

Explanation of Solution

Given info:

  $z=\frac{\cos \pi +i\sin \pi }{\cos \left(\frac{\pi }{3}\right)+i\sin \left(\frac{\pi }{3}\right)}$

Formula used:

Let $z_1=r_1\left(\cos {\theta }_1+i\sin {\theta }_1\right)\text{ and }{z}_2=r_2\left(\cos {\theta }_2+i\sin {\theta }_2\right)$ we have

  $\begin{array}{l}{z}_1{z}_2=r_1{r}_2\left(\cos \left({\theta }_1+{\theta }_2\right)+i\sin \left({\theta }_1+{\theta }_2\right)\right)\\ \frac{z_1}{z_2}=\frac{r_1}{r_2}\left(\cos \left({\theta }_1-{\theta }_2\right)+i\sin \left({\theta }_1-{\theta }_2\right)\right)\end{array}$

Calculation:

We have

  $\begin{array}{l}z=\frac{\cos \pi +i\sin \pi }{\cos \left(\frac{\pi }{3}\right)+i\sin \left(\frac{\pi }{3}\right)}\\ z_1=r_1\left(\cos {\theta }_1+i\sin {\theta }_1\right)\text{=}\cos \pi +i\sin \pi \\ z_2=r_2\left(\cos {\theta }_2+i\sin {\theta }_2\right)=\cos \left(\frac{\pi }{3}\right)+i\sin \left(\frac{\pi }{3}\right)\\ z=\frac{z_1}{z_2}=\frac{r_1}{r_2}\left(\cos \left({\theta }_1-{\theta }_2\right)+i\sin \left({\theta }_1-{\theta }_2\right)\right)\\ z=\frac{1}{1}\left(\cos \left(\pi -\frac{\pi }{3}\right)+i\sin \left(\pi -\frac{\pi }{3}\right)\right)\\ z=\cos \left(\frac{2\pi }{3}\right)+i\sin \left(\frac{2\pi }{3}\right)\end{array}$

Thequotient of the complex numbers in trigonometric form is $z=\cos \left(\frac{2\pi }{3}\right)+i\sin \left(\frac{2\pi }{3}\right)$

Conclusion:

Thus,thequotient of the complex numbers in trigonometric form is $z=\cos \left(\frac{2\pi }{3}\right)+i\sin \left(\frac{2\pi }{3}\right)$