Chapter 9.7, Problem 24E

To determine

To find the product and express the results in rectangular form.

Answer to Problem 24E

The value of product is $12\left(\cos {450}^{\circ }+i\sin {450}^{\circ }\right)$ and the rectangular form of the product is $12i$

Explanation of Solution

Given information:

The equation is

  $\left(2-2i\right)\times \left(-3+3i\right)$

Calculation:

The product can be obtained as

  $\left(2-2i\right)\times \left(-3+3i\right)$

First convert the rectangular form of the given complex number to polar form

  $\left(2-2i\right)$

First find the absolute value of r

  $\begin{array}{l}r=\sqrt{{\left(2\right)}^2+{\left(-2\right)}^2}\\ r=\sqrt{4+4}\\ r=\sqrt{8}\\ r=\sqrt{2\times 2\times 2}\\ r=2\sqrt{2}\left[\text{since, }\sqrt{2\times 2}=2\right]\end{array}$

Now find the value of $\theta$

  $\begin{array}{l}\theta =\arctan \left(\frac{2}{2}\right)\\ \theta =\arctan \left(1\right)\\ \theta ={45}^{\circ }\end{array}$

The complex number is in the fourth coordinate

So,

  $\begin{array}{l}\theta ={360}^{\circ }-{45}^{\circ }\\ \theta ={315}^{\circ }\end{array}$

Therefore, the polar form is

  $2\sqrt{2}\left(\cos {315}^{\circ }+i\sin {315}^{\circ }\right)$

Again convert the rectangular form of the given complex number to polar form

  $\left(-3+3i\right)$

First find the absolute value of r

  $\begin{array}{l}r=\sqrt{{\left(-3\right)}^2+{\left(3\right)}^2}\\ r=\sqrt{9+9}\\ r=\sqrt{18}\\ r=\sqrt{2\times 3\times 3}\\ r=3\sqrt{2}\left[\text{since, }\sqrt{3\times 3}=3\right]\end{array}$

Now find the value of $\theta$

  $\begin{array}{l}\theta =\arctan \left(\frac{3}{3}\right)\\ \theta =\arctan \left(1\right)\\ \theta ={45}^{\circ }\end{array}$

The complex number is in the second coordinate

So,

  $\begin{array}{l}\theta ={180}^{\circ }-{45}^{\circ }\\ \theta ={135}^{\circ }\end{array}$

Therefore, the polar form is

  $3\sqrt{2}\left(\cos {135}^{\circ }+i\sin {135}^{\circ }\right)$

Find the modulus and amplitude of the product.

  $\begin{array}{l}r=r_1{r}_2\left[r=\text{modulus}\right]\\ r\text{ =}\left(2\sqrt{2}\right)\left(3\sqrt{2}\right)\left[\text{since, }{r}_1=2\sqrt{2}\text{ and }{r}_2=3\sqrt{2}\right]\\ r=12\end{array}$

And

  $\begin{array}{l}\theta ={\theta }_1+{\theta }_2\left[\theta =\text{amplitude}\right]\\ \theta {\text{ =315}}^{\circ }+{135}^{\circ }\left[\text{since, }{\theta }_1={\text{315}}^{\circ }\text{ and }{\theta }_2={135}^{\circ }\right]\\ \theta {\text{ =450}}^{\circ }\end{array}$

The product is $12\left(\cos {450}^{\circ }+i\sin {450}^{\circ }\right)$

Now find the rectangular form of the product.

  $\begin{array}{l}12\left(\cos {450}^{\circ }+i\sin {450}^{\circ }\right)=12\left(\left(0\right)+i\left(1\right)\right)\left[\text{since, }\cos {450}^{\circ }=0\text{ and }\sin {450}^{\circ }=1\right]\\ =12\left(0+i\right)\\ =12i\end{array}$

The rectangular form of the product is $12i$

Hence,

The value of product is $12\left(\cos {450}^{\circ }+i\sin {450}^{\circ }\right)$ and the rectangular form of the product is $12i$