Chapter 6.5, Problem 69E

To determine

To find:

The multiplication of complex numbers in the form of trigonometric

Answer to Problem 69E

The complex number of multiplication is $z_1{z}_2=\left[\cos \left({490}^0\right)+i\sin \left({490}^0\right)\right]$

Explanation of Solution

Given:

The form is

  $\left[\left(\cos {290}^0+i\sin {290}^0\right)\right]\left[\left(\cos {200}^0+i\sin {200}^0\right)\right]$

Concept used:

Multiplication and division of Complex Number:- Let two complex numbers are

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

  $\begin{array}{l}{z}_1{z}_2=r_1{r}_2\left[\cos \left({\theta }_1+{\theta }_2\right)+i\sin \left({\theta }_1+{\theta }_2\right)\right]\\ \frac{z_1}{z_2}=\frac{r_1}{r_2}\left[\cos \left({\theta }_1-{\theta }_2\right)+i\sin \left({\theta }_1-{\theta }_2\right)\right];z_2\ne 0\end{array}$

Calculation:

The complex number of multiplication

  $\left[\left(\cos {290}^0+i\sin {290}^0\right)\right]\left[\left(\cos {200}^0+i\sin {200}^0\right)\right]\mathrm{..........}\left(1\right)$

Using the formula of multiplication in equation (1)

  $\begin{array}{l}{z}_1{z}_2=r_1{r}_2\left[\cos \left({\theta }_1+{\theta }_2\right)+i\sin \left({\theta }_1+{\theta }_2\right)\right]\\ z_1{z}_2=1\times 1\left[\cos \left({290}^0+{200}^0\right)+i\sin \left({290}^0+{200}^0\right)\right]\\ z_1{z}_2=\left[\cos \left({490}^0\right)+i\sin \left({490}^0\right)\right]\end{array}$