Chapter 9.7, Problem 14E

To determine

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

Answer to Problem 14E

The value of product is $18\left(\cos \frac{\pi }{2}+i\sin \frac{\pi }{2}\right)$ and the rectangular form of the product is $\text{18}i$

Explanation of Solution

Given information:

The equation is

  $6\left[\cos \left(-\frac{\pi }{3}\right)+i\sin \left(-\frac{\pi }{3}\right)\right]\times 3\left(\cos \frac{5\pi }{6}+i\sin \frac{5\pi }{6}\right)$

Calculation:

The product can be obtained as

  $6\left[\cos \left(-\frac{\pi }{3}\right)+i\sin \left(-\frac{\pi }{3}\right)\right]\times 3\left(\cos \frac{5\pi }{6}+i\sin \frac{5\pi }{6}\right)$

First find the modulus and amplitude of the product.

  $\begin{array}{l}r=r_1{r}_2\left[r=\text{modulus}\right]\\ r\text{ =}\left(6\right)\left(3\right)\left[\text{since, }{r}_1=6\text{ and }{r}_2=3\right]\\ r=\text{18}\end{array}$

And

  $\begin{array}{l}\theta ={\theta }_1+{\theta }_2\left[\theta =\text{amplitude}\right]\\ \theta \text{ =}-\frac{\pi }{3}+\frac{5\pi }{6}\left[\text{since, }{\theta }_1=-\frac{\pi }{3}\text{ and }{\theta }_2=\frac{5\pi }{6}\right]\\ \theta \text{ =}\frac{-2\pi +5\pi }{6}\\ \theta =\frac{3\pi }{6}\\ \theta =\frac{\pi }{2}\left[\text{divide numerator and denominator by 3}\right]\end{array}$

The product is $18\left(\cos \frac{\pi }{2}+i\sin \frac{\pi }{2}\right)$

Now find the rectangular form of the product.

  $\begin{array}{l}18\left(\cos \frac{\pi }{2}+i\sin \frac{\pi }{2}\right)=18\left(\left(0\right)+i\left(1\right)\right)\left[\text{since, }\cos \frac{\pi }{2}=0\text{ and }\sin \frac{\pi }{2}=1\right]\\ \text{ =18}\left(i\right)\\ \text{ =18}i\end{array}$

The rectangular form of the product is $\text{18}i$

Hence,

The value of product is $18\left(\cos \frac{\pi }{2}+i\sin \frac{\pi }{2}\right)$ and the rectangular form of the product is $\text{18}i$