Chapter 6, Problem 120RE

a.

To determine

The two complex numbers in trigonometric form.

Answer to Problem 120RE

  $z_1=3\sqrt{2}\left(\cos \frac{5\pi }{4}+i\sin \frac{5\pi }{4}\right)$ and $z_2=4\left(\cos \frac{\pi }{6}+i\sin \frac{\pi }{6}\right)$ .

Explanation of Solution

Given information:

Write the two complex numbers in trigonometric form$z_1=-3(1+i),z_2=2(\sqrt{3}+i)$

Calculation:

From the given information, write the two complex numbers in trigonometric form are as shown below:

  $\begin{array}{l}{z}_1=-3(1+i)\\ z_2=2(\sqrt{3}+i)\end{array}$

To find $r,\theta$ so that we can write:

  $z=r(\cos \theta +i\sin \theta )$

Firstly find the value of r , by using the formula:

  $r=\sqrt{a^2+b^2},\text{where }z=a+bi$

  $\begin{array}{l}{z}_1=-3-3i=a_1+b_{1}i\\ a_1=-3\\ b_1=-3\\ r_1=\sqrt{{(-3)}^2+{(-3)}^2}=18=3\sqrt{2}\\ z_2=2\sqrt{3}+2i=a_2+b_{2}i\\ a_2=2\sqrt{3}\end{array}$

  $\begin{array}{l}{b}_2=2\\ r_2=\sqrt{{(2\sqrt{3})}^2+2^2}=\sqrt{16}=4\end{array}$

Now find

  $\theta$ by computing

  $\tan \theta$ using the formula:

  $\begin{array}{l}\tan \theta =\frac{b}{a}\\ \tan {\theta }_1=\frac{-3}{-3}=1\end{array}$

  ${\theta }_1=\frac{5\pi }{4}$ (Because

  ${\theta }_1$ is in quadrant III)

  ${\theta }_2=\frac{\pi }{6}$ (Because

  ${\theta }_2$ is in quadrant I)

Now write the trigonometric form of the complex numbers

  $z_1,z_2$ :

  $\begin{array}{l}{z}_1=3\sqrt{2}\left(\cos \frac{5\pi }{4}+i\sin \frac{5\pi }{4}\right)\\ z_2=4\left(\cos \frac{\pi }{6}+i\sin \frac{\pi }{6}\right)\end{array}$

Thus, two complex numbers in trigonometric forms are $z_1=3\sqrt{2}\left(\cos \frac{5\pi }{4}+i\sin \frac{5\pi }{4}\right)$ and $z_2=4\left(\cos \frac{\pi }{6}+i\sin \frac{\pi }{6}\right)$ .

b.

To determine

The complex numbers by using trigonometric forms.

Answer to Problem 120RE

  $z_1{z}_2=12\sqrt{2}\left(\cos \frac{17\pi }{12}+i\sin \frac{17\pi }{12}\right)$ and $\frac{z_1}{z_2}=\frac{3\sqrt{2}}{4}\left(\cos \left(\frac{13\pi }{6}\right)+i\sin \left(\frac{13\pi }{6}\right)\right)$ .

Explanation of Solution

Given information:

Use the trigonometric forms to find

  $z_1{z}_2$ and $\frac{z_1}{z_2}$ where

  $z_2\ne 0$ .

  $z_1=-3(1+i),z_2=2(\sqrt{3}+i)$

Calculation:

From the given information, find

  $z_1{z}_2$ by using the formula:

  $\begin{array}{l}{z}_1{z}_2=r_1{r}_2[\cos ({\theta }_1+{\theta }_2)+i\sin ({\theta }_1+{\theta }_2)]\\ z_1{z}_2=3\sqrt{2}\cdot 4[\cos (\frac{5\pi }{4}+\frac{\pi }{6})+i\sin (\frac{5\pi }{4}+\frac{\pi }{6})]\\ z_1{z}_2=12\sqrt{2}\left(\cos \frac{17\pi }{12}+i\sin \frac{17\pi }{12}\right)\end{array}$

Now, find

  $\frac{z_1}{z_2}$ by using the formula:

  $\begin{array}{l}\frac{z_1}{z_2}=\frac{r_1}{r_2}[\cos \left({\theta }_1-{\theta }_2\right)+i\sin \left({\theta }_1-{\theta }_2\right)]\\ \frac{z_1}{z_2}=\frac{3\sqrt{2}}{4}[\cos \left(\frac{5\pi }{4}-\frac{\pi }{6}\right)+i\sin \left(\frac{5\pi }{4}-\frac{\pi }{6}\right)]\\ \frac{z_1}{z_2}=\frac{3\sqrt{2}}{4}\left(\cos \left(\frac{13\pi }{6}\right)+i\sin \left(\frac{13\pi }{6}\right)\right)\end{array}$

Thus, two complex numbers in trigonometric forms are $z_1{z}_2=12\sqrt{2}\left(\cos \frac{17\pi }{12}+i\sin \frac{17\pi }{12}\right)$ and $\frac{z_1}{z_2}=\frac{3\sqrt{2}}{4}\left(\cos \left(\frac{13\pi }{6}\right)+i\sin \left(\frac{13\pi }{6}\right)\right)$ .