Chapter 6.6, Problem 68E

(a)

To determine

To calculate:

The trigonometric form of following complex number:

  $-4\left(1+i\right)$

Answer to Problem 68E

The trigonometric from of following complex number is $4\sqrt{2}\left(\cos 0+i\sin 0\right)\left(\cos \frac{\pi }{4}+i\sin \frac{\pi }{4}\right)$ .

Explanation of Solution

Given information:

  $-4\left(1+i\right)$

Calculation:

Consider,

  $\begin{array}{l}{z}_1=\left(-4+0i\right)\\ z_2=\left(1+i\right)\end{array}$

For $z_1=\left(-4+0i\right)$ :

Compare $z_1=\left(-4+0i\right)$ with $z=a+bi$ :

  $\begin{array}{l}a=-4\\ b=0\end{array}$

The modulus is calculated as:

  $\begin{array}{l}r=\sqrt{a^2+b^2}\\ r=\sqrt{{\left(-4\right)}^2+{\left(0\right)}^2}\\ r=\sqrt{16+0}\\ r=4\end{array}$

The argument is calculated as:

  $\begin{array}{l}\theta ={\tan }^{-1}\left(\frac{b}{a}\right)\\ \theta ={\tan }^{-1}\left(\frac{0}{-4}\right)\\ \theta ={\tan }^{-1}\left(0\right)\\ \theta =0\end{array}$

The trigonometric form of the complex number is calculated as:

  $\begin{array}{l}{z}_1=r\left(\cos \theta +i\sin \theta \right)\\ z_1=4\left(\cos 0+i\sin 0\right)\end{array}$

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:

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

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:

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

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

  $\begin{array}{l}-4\left(1+i\right)=\left[4\left(\cos 0+i\sin 0\right)\right]\left[\sqrt{2}\left(\cos \frac{\pi }{4}+i\sin \frac{\pi }{4}\right)\right]\\ -4\left(1+i\right)=4\sqrt{2}\left(\cos 0+i\sin 0\right)\left(\cos \frac{\pi }{4}+i\sin \frac{\pi }{4}\right)\end{array}$

(b)

To determine

To calculate:

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

Answer to Problem 68E

The product of complex number of the trigonometric form is $\left[4\left(\cos 0+i\sin 0\right)\right]\left[\sqrt{2}\left(\cos \frac{\pi }{4}+i\sin \frac{\pi }{4}\right)\right]=4\sqrt{2}\left[\cos \frac{\pi }{4}+i\sin \frac{\pi }{4}\right]$

Explanation of Solution

Given information:

  $\left[4\left(\cos 0+i\sin 0\right)\right]\left[\sqrt{2}\left(\cos \frac{\pi }{4}+i\sin \frac{\pi }{4}\right)\right]$

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:

  $\left[4\left(\cos 0+i\sin 0\right)\right]\left[\sqrt{2}\left(\cos \frac{\pi }{4}+i\sin \frac{\pi }{4}\right)\right]$

On comparing it with formula:

  $r_1=4,r_2=\sqrt{2},{\theta }_1=0,{\theta }_2=\frac{\pi }{4}$

Now,

  $\begin{array}{l}\left[4\left(\cos 0+i\sin 0\right)\right]\left[\sqrt{2}\left(\cos \frac{\pi }{4}+i\sin \frac{\pi }{4}\right)\right]=4\cdot \sqrt{2}\left[\cos \left(0+\frac{\pi }{4}\right)+i\sin \left(0+\frac{\pi }{4}\right)\right]\\ \left[4\left(\cos 0+i\sin 0\right)\right]\left[\sqrt{2}\left(\cos \frac{\pi }{4}+i\sin \frac{\pi }{4}\right)\right]=4\sqrt{2}\left[\cos \frac{\pi }{4}+i\sin \frac{\pi }{4}\right]\end{array}$

(c)

To determine

To calculate:

The product of complex number of the standard form

Answer to Problem 68E

The product of complex number of the standard form is $-4-4i$ .

Explanation of Solution

Given information:

  $-4\left(1+i\right)$

Calculation:

Know that:

The multiplication of given two vectors $\left(-4-0i\right)$ and $\left(1+i\right)$ is calculated as:

  $-4\left(1+i\right)=-4-4i$