Chapter 9.7, Problem 8CFU

To determine

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

Answer to Problem 8CFU

The product of equation $\left(2+2\sqrt{3}i\right).\left(-3+\sqrt{3}i\right)$ is $12\sqrt{3}\left(\cos \frac{\pi }{6}+i\sin \frac{\pi }{6}\right)$

The rectangular form of the product $12\sqrt{3}\left(\cos \frac{\pi }{6}+i\sin \frac{\pi }{6}\right)$ is $18+6\sqrt{3}i$

Explanation of Solution

Given information

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

Formula used:for product

  $\begin{array}{l}{r}_1\left(\cos {\theta }_1+i\sin {\theta }_1\right).r_2\left(\cos {\theta }_2+i\sin {\theta }_2\right)=r_1{r}_2\left(\cos \left({\theta }_1+{\theta }_2\right)+i\sin \left({\theta }_1+{\theta }_2\right)\right),\\ \left[\text{for modulus}\right]\\ r=r_1{r}_{2,}\\ \left[\text{for amplitude}\right]\\ \theta ={\theta }_1+{\theta }_2\end{array}$

Calculation:

Express each complex number in polar forms.

  $\begin{array}{l}\left(2+2\sqrt{3}i\right).\left(-3+\sqrt{3}i\right)\\ =4\left(\cos \frac{\pi }{3}+i\sin \frac{\pi }{3}\right).3\sqrt{3}\left[\cos \left(-\frac{\pi }{6}\right)+i\sin \left(-\frac{\pi }{6}\right)\right]\\ \left[r_1=\sqrt{2^2+{\left(2\sqrt{3}\right)}^2}=\sqrt{16}=4,r_2=\sqrt{{\left(-3\right)}^2+{\left(\sqrt{3}\right)}^2}=\sqrt{12}=3\sqrt{3}\right]\\ \left[{\theta }_1=\arctan \frac{2\sqrt{3}}{2}=\frac{\pi }{3},{\theta }_2=\arctan \frac{-\sqrt{3}}{3}=-\frac{\pi }{6}\right]\end{array}$

Firstly,finding the modulus and amplitude of the product.

  $\begin{array}{l}r=r_{1r}{r}_2=4\times 3\sqrt{3}=12\sqrt{3}\\ \theta ={\theta }_1+{\theta }_2\\ \theta =\frac{\pi }{3}-\frac{\pi }{6}\\ \theta =\frac{\pi }{6}\end{array}$

The product of $4\left(\cos \frac{\pi }{3}+i\sin \frac{\pi }{3}\right).3\sqrt{3}\left[\cos \left(-\frac{\pi }{6}\right)+i\sin \left(-\frac{\pi }{6}\right)\right]$ is $12\sqrt{3}\left(\cos \frac{\pi }{6}+i\sin \frac{\pi }{6}\right)$

Now, find the rectangular form of the product is $\begin{array}{l}12\sqrt{3}\left(\cos \frac{\pi }{6}+i\sin \frac{\pi }{6}\right)=12\sqrt{3}\left(\frac{\sqrt{3}}{2}+i\frac{1}{2}\right)\left[\cos \frac{\pi }{6}=\frac{\sqrt{3}}{2},\sin \frac{\pi }{6}=\frac{1}{2}\right]\\ =\left(\frac{12\sqrt{3}.\sqrt{3}}{2}+i\frac{12\sqrt{3}}{2}\right)=18+6\sqrt{3}i\end{array}$

The rectangular form of the product $12\sqrt{3}\left(\cos \frac{\pi }{6}+i\sin \frac{\pi }{6}\right)$ is $18+6\sqrt{3}i$