Chapter 6.6, Problem 61E

(a)

To determine

To calculate:

The trigonometric form of following complex number:

  $\left(2+2i\right)\left(1-i\right)$

Answer to Problem 61E

The trigonometric from of given complex number is

  $4\left(\cos \frac{\pi }{4}+i\sin \frac{\pi }{4}\right)\left(\cos \left(-\frac{\pi }{4}\right)+i\sin \left(-\frac{\pi }{4}\right)\right)$ .

Explanation of Solution

Given information:

  $\left(2+2i\right)\left(1-i\right)$

Calculation:

Consider,

  $\begin{array}{l}{z}_1=\left(2+2i\right)\\ z_2=\left(1-i\right)\end{array}$

For $z_1=\left(2+2i\right)$ :

Compare $z_1=\left(2+2i\right)$ with $z=a+bi$ :

  $\begin{array}{l}a=2\\ b=2\end{array}$

The modulus is calculated as:

  $\begin{array}{l}r=\sqrt{a^2+b^2}\\ r=\sqrt{{\left(2\right)}^2+{\left(2\right)}^2}\\ r=\sqrt{4+4}\\ r=2\sqrt{2}\end{array}$

The argument is calculated as:

  $\begin{array}{l}\theta ={\tan }^{-1}\left(\frac{b}{a}\right)\\ \theta ={\tan }^{-1}\left(\frac{2}{2}\right)\\ \theta ={\tan }^{-1}\left(1\right)\\ \theta =\frac{\pi }{4}\end{array}$

The trigonometric form of the complex number is calculated as:

  $\begin{array}{l}{z}_1=r\left(\cos \theta +i\sin \theta \right)\\ z_1=2\sqrt{2}\left(\cos \frac{\pi }{4}+i\sin \frac{\pi }{4}\right)\end{array}$

For $z_2=\left(1-i\right)$ :

Compare $z_2=\left(1-i\right)$ with $z=a+bi$ :

  $\begin{array}{l}a=1\\ b=-1\end{array}$

The modulus is calculated as:

  $\begin{array}{l}r=\sqrt{a^2+b^2}\\ r=\sqrt{{\left(1\right)}^2+{\left(-1\right)}^2}\\ r=\sqrt{2}\\ r=\sqrt{2}\end{array}$

The argument is calculated as:

  $\begin{array}{l}\theta ={\tan }^{-1}\left(\frac{b}{a}\right)\\ \theta ={\tan }^{-1}\left(\frac{-1}{1}\right)\\ \theta ={\tan }^{-1}\left(-1\right)\\ \theta =-\frac{\pi }{4}\end{array}$

The trigonometric form of the complex number is calculated as:

  $\begin{array}{l}{z}_2=r\left(\cos \theta +i\sin \theta \right)\\ z_2=\sqrt{2}\left(\cos \left(-\frac{\pi }{4}\right)+i\sin \left(-\frac{\pi }{4}\right)\right)\end{array}$

Therefore, trigonometric form for the given complex number is calculated as:

  $\begin{array}{l}\left(2+2i\right)\left(1-i\right)=\left[2\sqrt{2}\left(\cos \frac{\pi }{4}+i\sin \frac{\pi }{4}\right)\right]\left[\sqrt{2}\left(\cos \left(-\frac{\pi }{4}\right)+i\sin \left(-\frac{\pi }{4}\right)\right)\right]\\ \left(2+2i\right)\left(1-i\right)=4\left(\cos \frac{\pi }{4}+i\sin \frac{\pi }{4}\right)\left(\cos \left(-\frac{\pi }{4}\right)+i\sin \left(-\frac{\pi }{4}\right)\right)\end{array}$

(b)

To determine

To calculate:

The product of complex number of the trigonometric form from subpart (a)

Answer to Problem 61E

The product of complex number of the trigonometric form is $\left[2\sqrt{2}\left(\cos \frac{\pi }{4}+i\sin \frac{\pi }{4}\right)\right]\left[\sqrt{2}\left(\cos \left(-\frac{\pi }{4}\right)+i\sin \left(-\frac{\pi }{4}\right)\right)\right]=4\left[\cos \left(\frac{\pi }{2}\right)+i\sin \left(-\frac{\pi }{2}\right)\right]$

Explanation of Solution

Given information:

  $\left[2\sqrt{2}\left(\cos \frac{\pi }{4}+i\sin \frac{\pi }{4}\right)\right]\left[\sqrt{2}\left(\cos \left(-\frac{\pi }{4}\right)+i\sin \left(-\frac{\pi }{4}\right)\right)\right]$

Formula used:

The multiplication rule for two complex numbers:

  $\begin{array}{l}{z}_1{z}_2=r_1\left(\cos {\theta }_1+i\sin {\theta }_1\right)\cdot r_2\left(\cos {\theta }_2+i\sin {\theta }_2\right)\\ z_1{z}_2=r_1\cdot r_2\left[\cos \left({\theta }_1+{\theta }_2\right)+i\sin \left({\theta }_1+{\theta }_2\right)\right]\end{array}$

Calculation:

The multiplication of complex number is calculated as:

  $\left[2\sqrt{2}\left(\cos \frac{\pi }{4}+i\sin \frac{\pi }{4}\right)\right]\left[\sqrt{2}\left(\cos \left(-\frac{\pi }{4}\right)+i\sin \left(-\frac{\pi }{4}\right)\right)\right]$

On comparing it with formula:

  $r_1=2\sqrt{2},r_2=\sqrt{2},{\theta }_1=\frac{\pi }{4},{\theta }_2=-\frac{\pi }{4}$

Now, $\begin{array}{l}\left[2\sqrt{2}\left(\cos \frac{\pi }{4}+i\sin \frac{\pi }{4}\right)\right]\left[\sqrt{2}\left(\cos \left(-\frac{\pi }{4}\right)+i\sin \left(-\frac{\pi }{4}\right)\right)\right]=2\sqrt{2}\cdot \sqrt{2}\left[\cos \left(\frac{\pi }{4}+\frac{\pi }{4}\right)+i\sin \left(\left(-\frac{\pi }{4}\right)+\left(-\frac{\pi }{4}\right)\right)\right]\\ \left[2\sqrt{2}\left(\cos \frac{\pi }{4}+i\sin \frac{\pi }{4}\right)\right]\left[\sqrt{2}\left(\cos \left(-\frac{\pi }{4}\right)+i\sin \left(-\frac{\pi }{4}\right)\right)\right]=4\left[\cos \left(\frac{\pi }{2}\right)+i\sin \left(-\frac{\pi }{2}\right)\right]\end{array}$

(c)

To determine

To calculate:

The product of complex number of the standard form

Answer to Problem 61E

The product of complex number of the standard form is $3$ .

Explanation of Solution

Given information:

  $\left(2+2i\right)\left(1-i\right)$

Calculation:

Know that:

The multiplication of given two vectors $\left(2+2i\right)$ and $\left(1-i\right)$ is calculated as:

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