Chapter 9.7, Problem 16E

To determine

To find the product and express the result in rectangular form.

Answer to Problem 16E

The value of product is $6\left(\cos {300}^{\circ }+i\sin {300}^{\circ }\right)$ and the rectangular form of the product is $\text{3}-i3\sqrt{3}$

Explanation of Solution

Given information:

The equation is

  $2\left(\cos {240}^{\circ }+i\sin {240}^{\circ }\right)\times 3\left(\cos {60}^{\circ }+i\sin {60}^{\circ }\right)$

Calculation:

The product can be obtained as

  $2\left(\cos {240}^{\circ }+i\sin {240}^{\circ }\right)\times 3\left(\cos {60}^{\circ }+i\sin {60}^{\circ }\right)$

First find the modulus and amplitude of the product.

  $\begin{array}{l}r=r_1{r}_2\left[r=\text{modulus}\right]\\ r\text{ =}\left(2\right)\left(3\right)\left[\text{since, }{r}_1=2\text{ and }{r}_2=3\right]\\ r=\text{6}\end{array}$

And

  $\begin{array}{l}\theta ={\theta }_1+{\theta }_2\left[\theta =\text{amplitude}\right]\\ \theta {\text{ =240}}^{\circ }+{60}^{\circ }\left[\text{since, }{\theta }_1={240}^{\circ }\text{ and }{\theta }_2={60}^{\circ }\right]\\ \theta {\text{ =300}}^{\circ }\end{array}$

The product is $6\left(\cos {300}^{\circ }+i\sin {300}^{\circ }\right)$

Now find the rectangular form of the product.

  $\begin{array}{l}6\left(\cos {300}^{\circ }+i\sin {300}^{\circ }\right)=6\left(\left(\frac{1}{2}\right)+i\left(-\frac{\sqrt{3}}{2}\right)\right)\left[\text{since, }\cos {300}^{\circ }=\frac{1}{2}\text{ and }\sin {300}^{\circ }=-\frac{\sqrt{3}}{2}\right]\\ \text{ =6}\left(\frac{1}{2}-i\frac{\sqrt{3}}{2}\right)\\ \text{ =3}-i3\sqrt{3}\end{array}$

The rectangular form of the product is $\text{3}-i3\sqrt{3}$

Hence,

The value of product is $6\left(\cos {300}^{\circ }+i\sin {300}^{\circ }\right)$ and the rectangular form of the product is $\text{3}-i3\sqrt{3}$