Chapter 9.7, Problem 3CFU

To determine

Tolist which operations with complex number is easier in rectangular form or polar form .

Answer to Problem 3CFU

The square of complex number in polar forms is $r^2{\left(\cos \theta +i\sin \theta \right)}^2$

Explanation of Solution

Given information

To square a complex number in polar form.

Calculation:

Two operations are product and quotient there to form complex number in rectangular form and in polar form.

To Expresscomplex number in polar forms.

First let product form of complex number in polar form and in rectangular form

Let two complex number will be $x_1+y_{1}i\text{ and }{x}_2+y_{2}i$

Now using product to get complex number in polar forms

  $\begin{array}{l}\left(x_1+y_{1}i\right)\text{. }\left(x_2+y_{2}i\right)=\\ =r_1\left(\cos {\theta }_1+i\sin {\theta }_1\right).r_2\left(\cos {\theta }_2+i\sin {\theta }_2\right)\left[x=r\cos \theta ,y=r\sin \theta \right]\\ =r_1{r}_2\left(\cos \left({\theta }_1+{\theta }_2\right)+i\sin \left({\theta }_1+{\theta }_2\right)\right),\\ \left[\text{for modulus}\right]\\ r=r_1{r}_{2,}\\ \left[\text{for amplitude}\right]\\ \theta ={\theta }_1+{\theta }_2\end{array}$

Complex number to polar form is $r_1{r}_2\left(\cos \left({\theta }_1+{\theta }_2\right)+i\sin \left({\theta }_1+{\theta }_2\right)\right)$

Now by putting the values of $\theta$ and r we get the equation in rectangular form example

Let equation $\frac{1}{2}\left(\cos \frac{\pi }{3}+i\sin \frac{\pi }{3}\right).6\left(\cos \frac{5\pi }{6}+i\sin \frac{5\pi }{6}\right)$

  $\begin{array}{l}r=r_{1r}{r}_2=\frac{1}{2}\times 6=3\\ \theta ={\theta }_1+{\theta }_2\\ \theta =\frac{\pi }{3}+\frac{5\pi }{6}\\ \theta =\frac{7\pi }{6}\end{array}$

The product of $\frac{1}{2}\left(\cos \frac{\pi }{3}+i\sin \frac{\pi }{3}\right).6\left(\cos \frac{5\pi }{6}+i\sin \frac{5\pi }{6}\right)$ is $3\left(\cos \frac{7\pi }{6}+i\sin \frac{7\pi }{6}\right)$

Now, find the rectangular form of the product is $\begin{array}{l}3\left(\cos \frac{7\pi }{6}+i\sin \frac{7\pi }{6}\right)=3\left(-\frac{\sqrt{3}}{2}-\frac{1}{2}i\right)\left[\cos \frac{7\pi }{6}=-\frac{\sqrt{3}}{2},\sin \frac{7\pi }{6}=-\frac{1}{2}\right]\\ =-\frac{3}{2}\left(\sqrt{3}+1i\right)=-\frac{3\sqrt{3}}{2}-\frac{3}{2}i\end{array}$

The rectangular form of the product $3\left(\cos \frac{7\pi }{6}+i\sin \frac{7\pi }{6}\right)$ is $-\frac{3\sqrt{3}}{2}-\frac{3}{2}i$

Now by using quotient to get complex number in polar forms

  $\begin{array}{l}\left(x_1+y_{1}i\right)÷\left(x_2+y_{2}i\right)=\\ =r_1\left(\cos {\theta }_1+i\sin {\theta }_1\right)÷r_2\left(\cos {\theta }_2+i\sin {\theta }_2\right)\left[x=r\cos \theta ,y=r\sin \theta \right]\\ =r_1\left(\cos {\theta }_1+i\sin {\theta }_1\right)÷r_2\left(\cos {\theta }_2+i\sin {\theta }_2\right)=\frac{r_1}{r_2}\left(\cos \left({\theta }_1-{\theta }_2\right)+i\sin \left({\theta }_1-{\theta }_2\right)\right),\\ \left[\text{for modulus}\right]\\ r=\frac{r_1}{r_2}\\ \left[\text{for amplitude}\right]\\ \theta ={\theta }_1-{\theta }_2\end{array}$

The complex number to polar form is $\frac{r_1}{r_2}\left(\cos \left({\theta }_1-{\theta }_2\right)+i\sin \left({\theta }_1-{\theta }_2\right)\right)$

Now by putting the values of $\theta$ and r we get the equation in rectangular form example

The quotient of $3\left(\cos \frac{\pi }{6}+i\sin \frac{\pi }{6}\right)÷4\left(\cos \frac{2\pi }{3}+i\sin \frac{2\pi }{3}\right)$ is $\frac{3}{4}\left(\cos \left(-\frac{\pi }{2}\right)+i\sin \left(-\frac{\pi }{2}\right)\right)$

Now, finding the rectangular form of the quotient is $\begin{array}{l}\frac{3}{4}\left(\cos \left(-\frac{\pi }{2}\right)+i\sin \left(-\frac{\pi }{2}\right)\right)=\frac{3}{4}\left(0-1i\right)\left[\cos \left(-\frac{\pi }{2}\right)=0,\sin \left(-\frac{\pi }{2}\right)=-1\right]\\ =0-\frac{3}{4}i\end{array}$

The rectangular form of the quotient $\frac{3}{4}\left(\cos \left(-\frac{\pi }{2}\right)+i\sin \left(-\frac{\pi }{2}\right)\right)$ is $0-\frac{3}{4}i$ .

Hence by examples we get that complex number to polar form is more easier than rectangular firm