Chapter 9.7, Problem 15E

To determine

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

Answer to Problem 15E

The value of quotient is $3\left(\cos \frac{11\pi }{6}+i\sin \frac{11\pi }{6}\right)$ and the rectangular form of the quotient is $\frac{3\sqrt{3}}{2}-\frac{3}{2}i$

Explanation of Solution

Given information:

The equation is

  $3\left(\cos \frac{7\pi }{3}+i\sin \frac{7\pi }{3}\right)÷\left(\cos \frac{\pi }{2}+i\sin \frac{\pi }{2}\right)$

Calculation:

The quotient can be obtained as

  $3\left(\cos \frac{7\pi }{3}+i\sin \frac{7\pi }{3}\right)÷\left(\cos \frac{\pi }{2}+i\sin \frac{\pi }{2}\right)$

First find the modulus and amplitude of the quotient.

  $\begin{array}{l}r=\frac{r_1}{r_2}\left[r=\text{modulus}\right]\\ r\text{ =}\frac{3}{1}\left[\text{since, }{r}_1=3\text{ and }{r}_2=1\right]\\ r=3\end{array}$

And

  $\begin{array}{l}\theta ={\theta }_1-{\theta }_2\left[\theta =\text{amplitude}\right]\\ \theta \text{ =}\frac{7\pi }{3}-\frac{\pi }{2}\left[\text{since, }{\theta }_1=\frac{7\pi }{3}\text{ and }{\theta }_2=\frac{\pi }{2}\right]\\ \theta \text{ =}\frac{14\pi -3\pi }{6}\\ \theta =\frac{11\pi }{6}\end{array}$

The quotient is $3\left(\cos \frac{11\pi }{6}+i\sin \frac{11\pi }{6}\right)$

Now find the rectangular form of the quotient.

  $\begin{array}{l}3\left(\cos \frac{11\pi }{6}+i\sin \frac{11\pi }{6}\right)=3\left(\frac{\sqrt{3}}{2}+i\left(-\frac{1}{2}\right)\right)\left[\text{since, }\cos \frac{11\pi }{6}=\frac{\sqrt{3}}{2}\text{ and }\sin \frac{11\pi }{6}=-\frac{1}{2}\right]\\ \text{ =}\frac{3\sqrt{3}}{2}-\frac{3}{2}i\end{array}$

The rectangular form of the quotient is $\frac{3\sqrt{3}}{2}-\frac{3}{2}i$

Hence,

The value of quotient is $3\left(\cos \frac{11\pi }{6}+i\sin \frac{11\pi }{6}\right)$ and the rectangular form of the quotient is $\frac{3\sqrt{3}}{2}-\frac{3}{2}i$