Chapter 9.7, Problem 6CFU

To determine

To find each product or quotient and express the result in rectangular form

Answer to Problem 6CFU

The quotient of equation $4\left(\cos \frac{9\pi }{4}+i\sin \frac{9\pi }{4}\right)÷2\left[\cos \left(-\frac{\pi }{2}\right)+i\sin \left(-\frac{\pi }{2}\right)\right]$ is $2\left(\cos \frac{11\pi }{4}+i\sin \frac{11\pi }{4}\right)$

The rectangular form of the quotient $2\left(\cos \frac{11\pi }{4}+i\sin \frac{11\pi }{4}\right)$ is $0-2i$

Explanation of Solution

Given information

The given form of equation is $4\left(\cos \frac{9\pi }{4}+i\sin \frac{9\pi }{4}\right)÷2\left[\cos \left(-\frac{\pi }{2}\right)+i\sin \left(-\frac{\pi }{2}\right)\right]$

Formula used:for quotient

  $\begin{array}{l}{r}_1\left(\cos {\theta }_1+i\sin {\theta }_1\right)÷r_2\left(\cos {\theta }_2+i\sin {\theta }_2\right)=\frac{r_1}{r_2}\left(\cos \left({\theta }_1-{\theta }_2\right)+i\sin \left({\theta }_1-{\theta }_2\right)\right),\\ \left[\text{for modulus}\right]\\ r=\frac{r_1}{r_2}\\ \left[\text{for amplitude}\right]\\ \theta ={\theta }_1-{\theta }_2\end{array}$

Calculation:

Firstly,finding the modulus and amplitude of the quotient.

  $\begin{array}{l}r=\frac{r_1}{r_2}=\frac{4}{2}=2\\ \theta ={\theta }_1-{\theta }_2\\ \theta =\frac{9\pi }{4}-\left(-\frac{\pi }{2}\right)=\frac{9\pi }{4}+\frac{\pi }{2}\\ \theta =\frac{9\pi +2\pi }{4}=\frac{11\pi }{4}\end{array}$

The quotient of $4\left(\cos \frac{9\pi }{4}+i\sin \frac{9\pi }{4}\right)÷2\left[\cos \left(-\frac{\pi }{2}\right)+i\sin \left(-\frac{\pi }{2}\right)\right]$ is $2\left(\cos \frac{11\pi }{4}+i\sin \frac{11\pi }{4}\right)$

Now, finding the rectangular form of the quotient is $\begin{array}{l}2\left(\cos \frac{11\pi }{4}+i\sin \frac{11\pi }{4}\right)=2\left(0-1i\right)\left[\cos \frac{11\pi }{4}=0,\sin \frac{11\pi }{4}=-1\right]\\ =0-2i\end{array}$

The rectangular form of the quotient $2\left(\cos \frac{11\pi }{4}+i\sin \frac{11\pi }{4}\right)$ is $0-2i$ .