Chapter 8.3, Problem 67E

To determine

To calculate: To write the complex numbers in polar form and then find the product and the quotient of the given complex numbers and express the answer in polar form

Answer to Problem 67E

Polar form of complex numbers are $z_1=20\left(\cos \pi +isin\pi \right)$ and $z_2=2\left(\cos \frac{\pi }{6}+isin\frac{\pi }{6}\right)$

Product is $z_1{z}_2=40\left(\cos \frac{7\pi }{6}+isin\frac{7\pi }{6}\right)$

Quotient is $\frac{z_1}{z_2}=10\left(\cos \frac{5\pi }{6}+isin\frac{5\pi }{6}\right)$

Explanation of Solution

Given information: Complex numbers are $z_1=-20$ and $z_2=\sqrt{3}+i$

Formula Used:

Complex number is a number that can be expressed in the form of $a+bi$ , where $a$ and $b$ are real numbers

  $z=a+bi$

Polar form of the complex number is given as

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

Product of two complex numbers $z_1=r_1\left(\cos {\theta }_1+i\sin {\theta }_1\right)$ and $z_2=r_2\left(\cos {\theta }_2+i\sin {\theta }_2\right)$ is given as

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

Quotient of two complex numbers $z_1=r_1\left(\cos {\theta }_1+i\sin {\theta }_1\right)$ and $z_2=r_2\left(\cos {\theta }_2+i\sin {\theta }_2\right)$ is given as

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

Calculation:

Complex number are given as

  $\begin{array}{l}{z}_1=-20\\ z_2=\sqrt{3}+i\end{array}$

Consider complex number:

  $z_1=-20$

Polar form of complex number is

  $z_1=r_1\left(\cos {\theta }_1+isin{\theta }_1\right)--(1)$

Calculating the value of $r_1$

  $\begin{array}{l}{r}_1=\sqrt{{\left(-20\right)}^2+{\left(0\right)}^2}\\ r_1=\sqrt{400}\\ r_1=20\end{array}$

Calculating the value of ${\theta }_1$

  $\begin{array}{l}\tan {\theta }_1=\frac{0}{-20}\\ {\theta }_1={\tan }^{-1}\frac{0}{-20}\\ {\theta }_1=\pi -0\\ {\theta }_1=\pi \end{array}$

Hence, complex number in polar form is

  $z_1=20\left(\cos \pi +isin\pi \right)$

Consider complex number:

  $z_2=\sqrt{3}+i$

Polar form of complex number is

  $z_2=r_2\left(\cos {\theta }_2+isin{\theta }_2\right)--(1)$

Calculating the value of $r_2$

  $\begin{array}{l}{r}_2=\sqrt{{\left(\sqrt{3}\right)}^2+{\left(1\right)}^2}\\ r_2=\sqrt{3+1}\\ r_2=\sqrt{4}\\ r_2=2\end{array}$

Calculating the value of ${\theta }_2$

  $\begin{array}{l}\tan {\theta }_2=\frac{1}{\sqrt{3}}\\ {\theta }_2={\tan }^{-1}\frac{1}{\sqrt{3}}\\ {\theta }_2=\frac{\pi }{6}\end{array}$

Hence, complex number in polar form is

  $z_2=2\left(\cos \frac{\pi }{6}+isin\frac{\pi }{6}\right)$

Thus, complex numbers in polar form are:

  $\begin{array}{l}{z}_1=20\left(\cos \pi +isin\pi \right)\\ z_2=2\left(\cos \frac{\pi }{6}+isin\frac{\pi }{6}\right)\end{array}$

Product of complex number is calculated as

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

Substituting the values, product is

  $\begin{array}{l}{z}_1{z}_2=20\cdot 2\left(\cos \left(\pi +\frac{\pi }{6}\right)+isin\left(\pi +\frac{\pi }{6}\right)\right)\\ z_1{z}_2=40\left(\cos \left(\frac{7\pi }{6}\right)+isin\left(\frac{7\pi }{6}\right)\right)\\ z_1{z}_2=40\left(\cos \frac{7\pi }{6}+isin\frac{7\pi }{6}\right)\end{array}$

Quotient of complex number is calculated as

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

Substituting the values, product is

  $\begin{array}{l}\frac{z_1}{z_2}=\frac{20}{2}\left(\cos \left(\pi -\frac{\pi }{6}\right)+isin\left(\pi -\frac{\pi }{6}\right)\right)\\ \frac{z_1}{z_2}=10\left(\cos \left(\frac{5\pi }{6}\right)+isin\left(\frac{5\pi }{6}\right)\right)\\ \frac{z_1}{z_2}=10\left(\cos \frac{5\pi }{6}+isin\frac{5\pi }{6}\right)\end{array}$

Conclusion:

Hence, Product is $z_1{z}_2=40\left(\cos \frac{7\pi }{6}+isin\frac{7\pi }{6}\right)$

Quotient is $\frac{z_1}{z_2}=10\left(\cos \frac{5\pi }{6}+isin\frac{5\pi }{6}\right)$