Chapter 9, Problem 53SGA

To determine

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

Answer to Problem 53SGA

  $\begin{array}{c}\text{Product}=12\left[\cos \left(\frac{2\pi }{3}\right)+i\sin \left(\frac{2\pi }{3}\right)\right]\\ \text{Rectangular form}=-6+6\sqrt{3}i\end{array}$

Explanation of Solution

Given information:

  $4\left(\cos \frac{\pi }{3}+i\sin \frac{\pi }{3}\right)\cdot 3\left(\cos \frac{\pi }{3}+i\sin \frac{\pi }{3}\right)$

Calculation:

The data extracted from the given complex number is as follows:

  $\begin{array}{c}{r}_1=4\\ r_2=3\\ {\theta }_1=\frac{\pi }{3}\\ {\theta }_2=\frac{\pi }{3}\end{array}$

The product can be evaluated as follows:

  $\begin{array}{c}\text{Product}=r_1\cdot r_2\left[\cos \left({\theta }_1+{\theta }_2\right)+i\sin \left({\theta }_1+{\theta }_2\right)\right]\\ =\left(4\times 3\right)\left[\cos \left(\frac{\pi }{3}+\frac{\pi }{3}\right)+i\sin \left(\frac{\pi }{3}+\frac{\pi }{3}\right)\right]\\ =12\left[\cos \left(\frac{2\pi }{3}\right)+i\sin \left(\frac{2\pi }{3}\right)\right]\end{array}$

Thus the product is $12\left[\cos \left(\frac{2\pi }{3}\right)+i\sin \left(\frac{2\pi }{3}\right)\right]$ .

Now solve the product to evaluate the rectangular form as shown below.

  $\begin{array}{c}12\left[\cos \left(\frac{2\pi }{3}\right)+i\sin \left(\frac{2\pi }{3}\right)\right]=12\left[-\frac{1}{2}+\frac{\sqrt{3}}{2}i\right]\\ =-6+6\sqrt{3}i\end{array}$

Thus the rectangular form is $-6+6\sqrt{3}i$ .