Chapter 6.5, Problem 77E

(A)

To determine

To find:

The trigonometric form of a complex numbers

Answer to Problem 77E

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

Explanation of Solution

Given:

The form is

  $\left(2-2i\right)\left(1+i\right)$

Concept used:

If $z=r\left(\cos \theta +i\sin \theta \right)$ is a complex number and $n$ is a positive integer

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

Calculation:

The power of a complex number

  $\left(2-2i\right)\left(1+i\right)\mathrm{..............}\left(1\right)$

Using the formula of DeMoviver’stheorem in equation (1)

  $z^n=\left[r^n\left(\cos n\theta +i\sin n\theta \right)\right]$

The absolute value of $z=\left(2-2i\right)\left(1+i\right)$ is

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

The argument $\theta$ is when $r=\left|2-2i\right|$

  $\begin{array}{l}\tan \theta =\frac{b}{a}=\frac{-2}{2}\\ \tan \theta =-1\\ \theta =\frac{7\pi }{4}\end{array}$

The complex form is

  $\begin{array}{l}z=\left[r\left(\cos \theta +i\sin \theta \right)\right]\\ z=\left[\left(2\sqrt{2}\right)\left(\cos \frac{7\pi }{4}+i\sin \frac{7\pi }{4}\right)\right]\end{array}$

The argument $\theta$ is when $r=\left|1+i\right|$

  $\begin{array}{l}\tan \theta =\frac{b}{a}=\frac{1}{1}\\ \tan \theta =1\\ \theta =\frac{\pi }{4}\end{array}$

The complex form is

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

(B)

To determine

To find:

The given operation using the trigonometric forms

Answer to Problem 77E

The complex number of multiplication is $4$

Explanation of Solution

Given:

The form is

  $\left(2-2i\right)\left(1+i\right)$

Concept used:

Multiplication and division of Complex Number:- Let two complex numbers are

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

  $\begin{array}{l}{z}_1{z}_2=r_1{r}_2\left[\cos \left({\theta }_1+{\theta }_2\right)+i\sin \left({\theta }_1+{\theta }_2\right)\right]\\ \frac{z_1}{z_2}=\frac{r_1}{r_2}\left[\cos \left({\theta }_1-{\theta }_2\right)+i\sin \left({\theta }_1-{\theta }_2\right)\right];z_2\ne 0\end{array}$

Calculation:

The trgnometric form of multiplication

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

(C)

To determine

To find:

The multiplication of complex numbers in the form of trigonometric

Explanation of Solution

Given:

The form is

  $\left[\left(2\sqrt{2}\right)\left(\cos \frac{7\pi }{4}+i\sin \frac{7\pi }{4}\right)\right]\left[\sqrt{2}\left(\cos \frac{\pi }{4}+i\sin \frac{\pi }{4}\right)\right]$

Concept used:

Multiplication and division of Complex Number:- Let two complex numbers are

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

  $z_1{z}_2=r_1{r}_2\left[\cos \left({\theta }_1+{\theta }_2\right)+i\sin \left({\theta }_1+{\theta }_2\right)\right]$

Calculation:

The complex number of multiplication

  $\left[\left(2\sqrt{2}\right)\left(\cos \frac{7\pi }{4}+i\sin \frac{7\pi }{4}\right)\right]\left[\sqrt{2}\left(\cos \frac{\pi }{4}+i\sin \frac{\pi }{4}\right)\right]\mathrm{..............}\left(1\right)$

Using the formula of multiplication in equation (1)

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