Chapter 9.7, Problem 26E

To determine

To use polar form tofind the product or quotient and express the result in rectangular form

Answer to Problem 26E

The quotient of equation $\left(\sqrt{3}-i\right)\left(-3+\sqrt{3}i\right)$ is $\frac{1}{2}\left[\cos \left(\frac{\pi }{6}\right)+i\sin \left(\frac{\pi }{6}\right)\right]$

The rectangular form of the quotient $\frac{1}{2}\left[\cos \left(\frac{\pi }{6}\right)+i\sin \left(\frac{\pi }{6}\right)\right]$ is $\frac{\sqrt{3}}{4}+\frac{1}{4}i$

Explanation of Solution

Given information

The given form of equation is $\left(\sqrt{3}-i\right)÷\left(2-2\sqrt{3}i\right)$

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:

Express each complex number in polar forms.

  $\begin{array}{l}\left(\sqrt{3}-i\right)÷\left(2-2\sqrt{3}i\right)\\ =2\left(\cos \left(-\frac{\pi }{6}\right)+i\sin \left(-\frac{\pi }{6}\right)\right)÷3\sqrt{3}\left[\cos \left(-\frac{\pi }{3}\right)+i\sin \left(-\frac{\pi }{3}\right)\right]\\ \left[r_1=\sqrt{{\left(\sqrt{3}\right)}^2+{\left(-1\right)}^2}=2,r_2=\sqrt{{\left(2\right)}^2+{\left(-2\sqrt{3}\right)}^2}=\sqrt{16}=4\right]\\ \left[{\theta }_1=\arctan \left(-\frac{1}{\sqrt{3}}\right)=-\frac{\pi }{6},{\theta }_2=\arctan \frac{-2\sqrt{3}}{2}=\arctan \left(-\sqrt{3}\right)=-\frac{\pi }{3}\right]\end{array}$

Firstly,finding the modulus and amplitude of the quotient.

  $\begin{array}{l}r=\frac{r_1}{r_2}=\frac{2}{4}=\frac{1}{2}\\ \theta ={\theta }_1-{\theta }_2\\ \theta =-\frac{\pi }{6}+\frac{\pi }{3}\\ \theta =\frac{-\pi +2\pi }{6}=\frac{\pi }{6}\end{array}$

The quotient of $2\left(\cos \left(-\frac{\pi }{6}\right)+i\sin \left(-\frac{\pi }{6}\right)\right)÷3\sqrt{3}\left[\cos \left(-\frac{\pi }{3}\right)+i\sin \left(-\frac{\pi }{3}\right)\right]$ is $\frac{1}{2}\left[\cos \left(\frac{\pi }{6}\right)+i\sin \left(\frac{\pi }{6}\right)\right]$

Now, find the rectangular form of the quotient is $\begin{array}{l}\frac{1}{2}\left[\cos \left(\frac{\pi }{6}\right)+i\sin \left(\frac{\pi }{6}\right)\right]=\frac{1}{2}\left(\frac{\sqrt{3}}{2}+\frac{1}{2}i\right)\left[\cos \left(-\frac{\pi }{2}\right)=\frac{\sqrt{3}}{2},\sin \left(-\frac{\pi }{2}\right)=\frac{1}{2}\right]\\ =\frac{\sqrt{3}}{4}+\frac{1}{4}i\end{array}$

The rectangular form of the quotient $\frac{1}{2}\left[\cos \left(\frac{\pi }{6}\right)+i\sin \left(\frac{\pi }{6}\right)\right]$ is $\frac{\sqrt{3}}{4}+\frac{1}{4}i$