Chapter 6, Problem 119RE

a.

To determine

The two complex numbers in trigonometric form.

Answer to Problem 119RE

  $z_1=4\left(\cos \frac{11\pi }{6}+i\sin \frac{11\pi }{6}\right)$ and $z_2=10\left(\cos \frac{3\pi }{2}+i\sin \frac{3\pi }{2}\right)$ .

Explanation of Solution

Given information:

Write the two complex numbers in trigonometric form$z_1=2\sqrt{3}-2i,z_2=-10i$

Calculation:

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

  $\begin{array}{l}{z}_1=2\sqrt{3}-2i\\ z_2=-10i\end{array}$

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:

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

  $\begin{array}{l}{z}_1=2\sqrt{3}-2i=a_1+b_{1}i\\ a_1=2\sqrt{3}\\ b_1=-2\\ r_1=\sqrt{{(2\sqrt{3})}^2+{(-2)}^2}=\sqrt{16}=4\\ z_2=-10i=a_2+b_{2}i\end{array}$

  $\begin{array}{l}{a}_2=0\\ b_2=-10\\ r_2=\sqrt{0^2+{(-10)}^2}=\sqrt{100}=10\end{array}$

Now find

  $\theta$ by computing

  $\tan \theta$ using the formula:

  $\begin{array}{l}\tan \theta =\frac{b}{a}\\ \tan {\theta }_1=\frac{-2}{2\sqrt{3}}=-\frac{\sqrt{3}}{3}\end{array}$

  ${\theta }_1=\frac{11\pi }{6}$ (Because

  ${\theta }_1$ is in quadrant IV)

  ${\theta }_1=\frac{3\pi }{2}$ (Because

  $z_2$ is on the negative part of Oy )

Now write the trigonometric form of the complex numbers

  $z_1,z_2$ :

  $\begin{array}{l}{z}_1=4\left(\cos \frac{11\pi }{6}+i\sin \frac{11\pi }{6}\right)\\ z_2=10\left(\cos \frac{3\pi }{2}+i\sin \frac{3\pi }{2}\right)\end{array}$

Thus, two complex numbers in trigonometric forms are $z_1=4\left(\cos \frac{11\pi }{6}+i\sin \frac{11\pi }{6}\right)$ and $z_2=10\left(\cos \frac{3\pi }{2}+i\sin \frac{3\pi }{2}\right)$ .

b.

To determine

The complex numbers by using trigonometric forms.

Answer to Problem 119RE

  $z_1{z}_2=40\left(\cos \frac{4\pi }{3}+i\sin \frac{4\pi }{3}\right)$ and $\frac{z_1}{z_2}=\frac{2}{5}[\cos \left(\frac{\pi }{3}\right)+i\sin \left(\frac{\pi }{3}\right)]$ .

Explanation of Solution

Given information:

Use the trigonometric forms to find

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

  $z_2\ne 0$ .

  $z_1=2\sqrt{3}-2i,z_2=-10i$

Calculation:

From the given information, find

  $z_1{z}_2$ by using the formula:

  $\begin{array}{l}{z}_1{z}_2=r_1{r}_2[\cos ({\theta }_1+{\theta }_2)+i\sin ({\theta }_1+{\theta }_2)]\\ z_1{z}_2=4\cdot 10[\cos (\frac{11\pi }{6}+\frac{3\pi }{2})+i\sin (\frac{11\pi }{6}+\frac{3\pi }{2})]\\ z_1{z}_2=40\left(\cos \frac{20\pi }{6}+i\sin \frac{20\pi }{6}\right)\\ z_1{z}_2=40\left(\cos \frac{10\pi }{3}+i\sin \frac{10\pi }{3}\right)\\ z_1{z}_2=40\left(\cos \frac{4\pi }{3}+i\sin \frac{4\pi }{3}\right)\end{array}$

Now, find

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

  $\begin{array}{l}\frac{z_1}{z_2}=\frac{r_1}{r_2}[\cos \left({\theta }_1-{\theta }_2\right)+i\sin \left({\theta }_1-{\theta }_2\right)]\\ \frac{z_1}{z_2}=\frac{4}{10}[\cos \left(\frac{11\pi }{6}-\frac{3\pi }{2}\right)+i\sin \left(\frac{11\pi }{6}-\frac{3\pi }{2}\right)]\\ \frac{z_1}{z_2}=\frac{2}{5}[\cos \left(\frac{2\pi }{6}\right)+i\sin \left(\frac{2\pi }{6}\right)]\\ \frac{z_1}{z_2}=\frac{2}{5}[\cos \left(\frac{\pi }{3}\right)+i\sin \left(\frac{\pi }{3}\right)]\end{array}$

Thus, two complex numbers in trigonometric forms are $z_1{z}_2=40\left(\cos \frac{4\pi }{3}+i\sin \frac{4\pi }{3}\right)$ and $\frac{z_1}{z_2}=\frac{2}{5}[\cos \left(\frac{\pi }{3}\right)+i\sin \left(\frac{\pi }{3}\right)]$ .