Chapter 6.5, Problem 87E

(A)

To determine

To find:

The trigonometric form of a complex numbers

Answer to Problem 87E

The complex form is $z=\left[5\left(\cos 0.93+i\sin 0.93\right)\right]$ , $z=\left[2\left(\cos \frac{5\pi }{3}+i\sin \frac{5\pi }{3}\right)\right]$

Explanation of Solution

Given:

The form is

  $\frac{3+4i}{1-\sqrt{3}i}$

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

  $\frac{3+4i}{1-\sqrt{3}i}\mathrm{..............}\left(1\right)$

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 $z=\frac{3+4i}{1-\sqrt{3}i}$ is

  $\begin{array}{l}r=\left|3+4i\right|=\sqrt{{\left(3\right)}^2+{\left(4\right)}^2}\\ r=5\\ r=\left|1-\sqrt{3}i\right|=\sqrt{{\left(1\right)}^2+{\left(-\sqrt{3}\right)}^2}\\ r=2\end{array}$

The argument $\theta$ is when $r=\left|3+4i\right|$

  $\begin{array}{l}\tan \theta =\frac{b}{a}=\frac{4}{3}\\ \tan \theta =\frac{4}{3}\\ \theta =0.93\end{array}$

The complex form is

  $\begin{array}{l}z=\left[r\left(\cos \theta +i\sin \theta \right)\right]\\ z=\left[5\left(\cos 0.93+i\sin 0.93\right)\right]\end{array}$

The argument $\theta$ is when $r=\left|1-\sqrt{3}i\right|$

  $\begin{array}{l}\tan \theta =\frac{b}{a}=\frac{-\sqrt{3}}{1}\\ \tan \theta =-\sqrt{3}\\ \theta =\frac{5\pi }{3}\end{array}$

The complex form is

  $\begin{array}{l}z=\left[r\left(\cos \theta +i\sin \theta \right)\right]\\ z=\left[2\left(\cos \frac{5\pi }{3}+i\sin \frac{5\pi }{3}\right)\right]\end{array}$

(B)

To determine

To find:

The given operation using the trigonometric forms

Answer to Problem 87E

The complex number of division is $\left(\frac{3}{4}-\sqrt{3}\right)+\left(\frac{3\sqrt{3}}{4}+1\right)i$

Explanation of Solution

Given:

The form is

  $\frac{3+4i}{1-\sqrt{3}i}$

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 dividing rule

  $\begin{array}{l}\frac{3+4i}{1-\sqrt{3}i}\times \left(\frac{1+\sqrt{3}i}{1+\sqrt{3}i}\right)\\ \frac{\left(3+4i\right)1+\sqrt{3}i}{{\left(1\right)}^2+{\left(\sqrt{3}i\right)}^2}\\ \left(\frac{3}{4}-\sqrt{3}\right)+\left(\frac{3\sqrt{3}}{4}+1\right)i\end{array}$

(C)

To determine

To find:

The dividationof complex numbers in the form of trigonometric

Explanation of Solution

Given:

The form is

  $\left[5\left(\cos 0.93+i\sin 0.93\right)\right]\left[2\left(\cos \frac{5\pi }{3}+i\sin \frac{5\pi }{3}\right)\right]$

Concept used:

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

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

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

Calculation:

The complex number of division

  $\left[5\left(\cos 0.93+i\sin 0.93\right)\right]\left[2\left(\cos \frac{5\pi }{3}+i\sin \frac{5\pi }{3}\right)\right]\mathrm{..............}\left(1\right)$

Using the formula of division in equation (1)

  $\begin{array}{l}\frac{z_1}{z_2}=\frac{r_1}{r_2}\left[\cos \left({\theta }_1-{\theta }_2\right)+i\sin \left({\theta }_1-{\theta }_2\right)\right];z_2\ne 0\\ \frac{z_1}{z_2}=\frac{5}{2}\left[\left(\cos 0.93+i\sin 0.93\right)\right]\left[\left(\cos \frac{5\pi }{3}+i\sin \frac{5\pi }{3}\right)\right]\\ z_1{z}_2=\left(\frac{3}{4}-\sqrt{3}\right)+\left(\frac{3\sqrt{3}}{4}+1\right)i\end{array}$