Chapter 12.7, Problem 48E

To determine

To find: the quotient.

Answer to Problem 48E

  $4\left(\cos \left(-\frac{\pi }{8}\right)+i\sin \left(-\frac{\pi }{8}\right)\right)$

Explanation of Solution

Given information:

  $16\left(\cos \frac{\pi }{8}+i\sin \frac{\pi }{8}\right)÷4\left(\cos \frac{\pi }{4}+i\sin \frac{\pi }{4}\right)$

Calculation:

Let $16\left(\cos \frac{\pi }{8}+i\sin \frac{\pi }{8}\right)÷4\left(\cos \frac{\pi }{4}+i\sin \frac{\pi }{4}\right)=r\left(\cos \theta +i\sin \theta \right)$ . Thus, the value of $r\text{ and }\theta$ can be evaluated as follows:

  $\begin{array}{c}r=\frac{16}{4}\\ =4\\ \theta =\frac{\pi }{8}-\frac{\pi }{4}\\ =-\frac{\pi }{8}\end{array}$

Expand the series as follows:

  $\begin{array}{c}16\left(\cos \frac{\pi }{8}+i\sin \frac{\pi }{8}\right)÷4\left(\cos \frac{\pi }{4}+i\sin \frac{\pi }{4}\right)=\frac{16\left(\cos \frac{\pi }{8}+i\sin \frac{\pi }{8}\right)}{4\left(\cos \frac{\pi }{4}+i\sin \frac{\pi }{4}\right)}\\ =4\left(\cos \left(-\frac{\pi }{8}\right)+i\sin \left(-\frac{\pi }{8}\right)\right)\end{array}$

Thus, the expanded form is $4\left(\cos \left(-\frac{\pi }{8}\right)+i\sin \left(-\frac{\pi }{8}\right)\right)$ .