Chapter 8.3, Problem 54E

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 54E

Product is $z_1{z}_2=\cos \pi +i\sin \pi$

Quotient is $\frac{z_1}{z_2}=\cos \frac{3\pi }{2}+i\sin \frac{3\pi }{2}$

Explanation of Solution

Given information: Complex numbers are $z_1=\cos \frac{\pi }{4}+i\sin \frac{\pi }{4}$ and $z_2=\cos \frac{3\pi }{4}+i\sin \frac{3\pi }{4}$

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

  $\begin{array}{l}{z}_1=\cos \frac{\pi }{4}+i\sin \frac{\pi }{4}\\ z_2=\cos \frac{3\pi }{4}+i\sin \frac{3\pi }{4}\end{array}$

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

  $\begin{array}{l}{z}_1{z}_2=1\cdot 1\left(\cos \left(\frac{\pi }{4}+\frac{3\pi }{4}\right)+i\sin \left(\frac{\pi }{4}+\frac{3\pi }{4}\right)\right)\\ z_1{z}_2=1\left(\cos \frac{4\pi }{4}+i\sin \frac{4\pi }{4}\right)\\ z_1{z}_2=\cos \pi +i\sin \pi \end{array}$

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

  $\begin{array}{l}\frac{z_1}{z_2}=\frac{1}{1}\left(\cos \left(\frac{\pi }{4}-\frac{3\pi }{4}\right)+i\sin \left(\frac{\pi }{4}-\frac{3\pi }{4}\right)\right)\\ \frac{z_1}{z_2}=1\left(\cos \left(-\frac{2\pi }{4}\right)+i\sin \left(-\frac{2\pi }{4}\right)\right)\\ \frac{z_1}{z_2}=\cos \left(-\frac{\pi }{2}\right)+i\sin \left(-\frac{\pi }{2}\right)\\ \frac{z_1}{z_2}=\cos \left(-\frac{\pi }{2}+2\pi \right)+i\sin \left(-\frac{\pi }{2}+2\pi \right)\\ \frac{z_1}{z_2}=\cos \frac{3\pi }{2}+i\sin \frac{3\pi }{2}\end{array}$

Conclusion:

Hence, Product is $z_1{z}_2=\cos \pi +i\sin \pi$

Quotient is $\frac{z_1}{z_2}=\cos \frac{3\pi }{2}+i\sin \frac{3\pi }{2}$