Chapter 6.5, Problem 57E

(a)

To determine

The trigonometric forms of thecomplex numbers.

Answer to Problem 57E

The trigonometric formis $(2+2i)(1-i)=\left[2\sqrt{2}\left(\cos \frac{\pi }{4}+i\sin \frac{\pi }{4}\right)\right]\left[\sqrt{2}\left(\cos \frac{7\pi }{4}+i\sin \frac{7\pi }{4}\right)\right]$

Explanation of Solution

Given info:

  $(2+2i)(1-i)$

Formula used:

The trigonometric form of the complex number $z=a+ib$ is

  $z=r\left(\cos \theta +i\sin \theta \right)$

Where $r=\sqrt{a^2+b^2}$ and $\theta ={\tan }^{-1}\left(\frac{b}{a}\right)$

Calculation:

We have

  $\begin{array}{l}z=(2+2i)(1-i)\\ \underset{\_}{2+2i}\\ a=2,b=2\\ r=\sqrt{a^2+b^2}=\sqrt{2^2+2^2}=2\sqrt{2}\\ \theta ={\tan }^{-1}\left(\frac{2}{2}\right)=\frac{\pi }{4}\\ z=r\left(\cos \theta +i\sin \theta \right)\\ 2+2i=2\sqrt{2}\left(\cos \frac{\pi }{4}+i\sin \frac{\pi }{4}\right)\\ \underset{\_}{1-i}\\ a=1,b=-1\\ r=\sqrt{a^2+b^2}=\sqrt{1^2+{\left(-1\right)}^2}=\sqrt{2}\\ \theta ={\tan }^{-1}\left(\frac{-1}{1}\right)=\frac{7\pi }{4}\\ z=r\left(\cos \theta +i\sin \theta \right)\\ 1-i=\sqrt{2}\left(\cos \frac{7\pi }{4}+i\sin \frac{7\pi }{4}\right)\\ z=(2+2i)(1-i)=\left[2\sqrt{2}\left(\cos \frac{\pi }{4}+i\sin \frac{\pi }{4}\right)\right]\left[\sqrt{2}\left(\cos \frac{7\pi }{4}+i\sin \frac{7\pi }{4}\right)\right]\end{array}$

The trigonometric form is $(2+2i)(1-i)=\left[2\sqrt{2}\left(\cos \frac{\pi }{4}+i\sin \frac{\pi }{4}\right)\right]\left[\sqrt{2}\left(\cos \frac{7\pi }{4}+i\sin \frac{7\pi }{4}\right)\right]$

Conclusion:

Thus,the trigonometric form is $(2+2i)(1-i)=\left[2\sqrt{2}\left(\cos \frac{\pi }{4}+i\sin \frac{\pi }{4}\right)\right]\left[\sqrt{2}\left(\cos \frac{7\pi }{4}+i\sin \frac{7\pi }{4}\right)\right]$

(b)

To determine

The product using the trigonometric forms.

Answer to Problem 57E

The product using the trigonometric form is $4$

Explanation of Solution

Given info:

  $(2+2i)(1-i)$

Formula used:

Let $z_1=r_1\left(\cos {\theta }_1+i\sin {\theta }_1\right)\text{ and }{z}_2=r_2\left(\cos {\theta }_2+i\sin {\theta }_2\right)$ we have

  $\begin{array}{l}{z}_1{z}_2=r_1{r}_2\left(\cos \left({\theta }_1+{\theta }_2\right)+i\sin \left({\theta }_1+{\theta }_2\right)\right)\\ \frac{z_1}{z_2}=\frac{r_1}{r_2}\left(\cos \left({\theta }_1-{\theta }_2\right)+i\sin \left({\theta }_1-{\theta }_2\right)\right)\end{array}$

Calculation:

From part a we have $(2+2i)(1-i)=\left[2\sqrt{2}\left(\cos \frac{\pi }{4}+i\sin \frac{\pi }{4}\right)\right]\left[\sqrt{2}\left(\cos \frac{7\pi }{4}+i\sin \frac{7\pi }{4}\right)\right]$

We have

  $\begin{array}{l}z=(2+2i)(1-i)=\left[2\sqrt{2}\left(\cos \frac{\pi }{4}+i\sin \frac{\pi }{4}\right)\right]\left[\sqrt{2}\left(\cos \frac{7\pi }{4}+i\sin \frac{7\pi }{4}\right)\right]\\ z_1=r_1\left(\cos {\theta }_1+i\sin {\theta }_1\right)\text{=}2\sqrt{2}\left(\cos \frac{\pi }{4}+i\sin \frac{\pi }{4}\right)\\ z_2=r_2\left(\cos {\theta }_2+i\sin {\theta }_2\right)=\sqrt{2}\left(\cos \frac{7\pi }{4}+i\sin \frac{7\pi }{4}\right)\\ z=z_1{z}_2=r_1{r}_2\left(\cos \left({\theta }_1+{\theta }_2\right)+i\sin \left({\theta }_1+{\theta }_2\right)\right)\\ z=2\sqrt{2}\times \sqrt{2}\left(\cos \left(\frac{\pi }{4}+\frac{7\pi }{4}\right)+i\sin \left(\frac{\pi }{4}+\frac{7\pi }{4}\right)\right)\\ z=4\left(\cos 2\pi +i\sin 2\pi \right)\\ z=4\end{array}$

The product using the trigonometric formis $4$

Conclusion:

Thus,the product using the trigonometric form is $4$

(c)

To determine

The product using the standard forms, and checkyour result with that of part (b).

Answer to Problem 57E

The product using standard form and trigonometric form are same.

Explanation of Solution

Given info:

  $(2+2i)(1-i)$

Formula used:

We have the results

  $i^2=-1$

Calculation:

Product from part b is $4$

The product using the standard forms is given by

  $\begin{array}{l}z=(2+2i)(1-i)\\ z=2-2i+2i-2i^2=2+2=4\end{array}$

Product using standard form and trigonometric form are same.

Conclusion:

Thus,the product using standard form and trigonometric form are same.