Chapter 6.6, Problem 40E

To determine

Product of the complex numbers in trigonometric form.

Answer to Problem 40E

Theproduct of the complex numbers in trigonometric form is $\frac{2}{5}\left(\cos {40}^0+i\sin {40}^0\right)$

Explanation of Solution

Given info:

  $z=\left[\frac{1}{2}\left(\cos {100}^0+i\sin {100}^0\right)\right]\left[\frac{4}{5}\left(\cos {300}^0+i\sin {300}^0\right)\right]$

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

  $\begin{array}{l}z=\left[\frac{1}{2}\left(\cos {100}^0+i\sin {100}^0\right)\right]\left[\frac{4}{5}\left(\cos {300}^0+i\sin {300}^0\right)\right]\\ z_1=r_1\left(\cos {\theta }_1+i\sin {\theta }_1\right)\text{=}\frac{1}{2}\left(\cos {100}^0+i\sin {100}^0\right)\\ z_2=r_2\left(\cos {\theta }_2+i\sin {\theta }_2\right)=\frac{4}{5}\left(\cos {300}^0+i\sin {300}^0\right)\\ z=z_1{z}_2=r_1{r}_2\left(\cos \left({\theta }_1+{\theta }_2\right)+i\sin \left({\theta }_1+{\theta }_2\right)\right)\\ z=\frac{1}{2}\times \frac{4}{5}\left(\cos \left({100}^0+{300}^0\right)+i\sin \left({100}^0+{300}^0\right)\right)\\ z=\frac{2}{5}\left(\cos {400}^0+i\sin {400}^0\right)\\ z=\frac{2}{5}\left(\cos {40}^0+i\sin {40}^0\right)\end{array}$

The product of the complex numbers in trigonometric form is $\frac{2}{5}\left(\cos {40}^0+i\sin {40}^0\right)$

Conclusion:

Thus,theproduct of the complex numbers in trigonometric formis $\frac{2}{5}\left(\cos {40}^0+i\sin {40}^0\right)$