Chapter 8.3, Problem 55E

To determine

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

Answer to Problem 55E

Product is $z_1{z}_2=15\left(\cos \frac{3\pi }{2}+i\sin \frac{3\pi }{2}\right)$

Quotient is $\frac{z_1}{z_2}=0.6\left(\cos \frac{7\pi }{6}-i\sin \frac{7\pi }{6}\right)$

Explanation of Solution

Given information: Complex numbers are $z_1=3\left(\cos \frac{\pi }{6}+i\sin \frac{\pi }{6}\right)$ and $z_2=5\left(\cos \frac{4\pi }{3}+i\sin \frac{4\pi }{3}\right)$

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

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

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

  $\begin{array}{l}{z}_1{z}_2=3\cdot 5\left(\cos \left(\frac{\pi }{6}+\frac{4\pi }{3}\right)+i\sin \left(\frac{\pi }{6}+\frac{4\pi }{3}\right)\right)\\ z_1{z}_2=15\left(\cos \left(\frac{\pi }{6}+\frac{8\pi }{6}\right)+i\sin \left(\frac{\pi }{6}+\frac{8\pi }{6}\right)\right)\\ z_1{z}_2=15\left(\cos \left(\frac{9\pi }{6}\right)+i\sin \left(\frac{9\pi }{6}\right)\right)\\ z_1{z}_2=15\left(\cos \frac{3\pi }{2}+i\sin \frac{3\pi }{2}\right)\end{array}$

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

  $\begin{array}{l}\frac{z_1}{z_2}=\frac{3}{5}\left(\cos \left(\frac{\pi }{6}-\frac{4\pi }{3}\right)+i\sin \left(\frac{\pi }{6}-\frac{4\pi }{3}\right)\right)\\ \frac{z_1}{z_2}=0.6\left(\cos \left(\frac{\pi }{6}-\frac{8\pi }{6}\right)+i\sin \left(\frac{\pi }{6}-\frac{8\pi }{6}\right)\right)\\ \frac{z_1}{z_2}=0.6\left(\cos \left(-\frac{7\pi }{6}\right)+i\sin \left(-\frac{7\pi }{6}\right)\right)\\ \frac{z_1}{z_2}=0.6\left(\cos \frac{7\pi }{6}-i\sin \frac{7\pi }{6}\right)\end{array}$

Conclusion:

Hence, Product is $z_1{z}_2=15\left(\cos \frac{3\pi }{2}+i\sin \frac{3\pi }{2}\right)$

Quotient is $\frac{z_1}{z_2}=0.6\left(\cos \frac{7\pi }{6}-i\sin \frac{7\pi }{6}\right)$