Chapter 6.5, Problem 62E

(a)

To determine

The trigonometric forms of the complex numbers.

Answer to Problem 62E

The trigonometric form is $\frac{(1+\sqrt{3}i)}{(6-3i)}=\frac{2\left(\cos {60}^0+i\sin {60}^0\right)}{3\sqrt{5}\left(\cos {333.43}^0+i\sin {333.43}^0\right)}$

Explanation of Solution

Given info:

  $\frac{(1+\sqrt{3}i)}{(6-3i)}$

Formula used:

The trigonometric form of the complex number $z=a+ib$ is

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

Where $r=\sqrt{a^2+b^2}$ and $\theta ={\tan }^{-1}\left(\frac{b}{a}\right)$

Calculation:

We have

  $\begin{array}{l}z=\frac{(1+\sqrt{3}i)}{(6-3i)}\\ \underset{\_}{1+\sqrt{3}i}\\ a=1,b=\sqrt{3}\\ r=\sqrt{a^2+b^2}=\sqrt{1^2+{\sqrt{3}}^2}=2\\ \theta ={\tan }^{-1}\left(\frac{\sqrt{3}}{1}\right)={60}^0\\ z=r\left(\cos \theta +i\sin \theta \right)\\ 1+\sqrt{3}i=2\left(\cos {60}^0+i\sin {60}^0\right)\\ \underset{\_}{(6-3i)}\\ a=6,b=-3\\ r=\sqrt{a^2+b^2}=\sqrt{6^2+{\left(-3\right)}^2}=3\sqrt{5}\\ \theta ={\tan }^{-1}\left(\frac{-3}{6}\right)={333.43}^0\\ z=r\left(\cos \theta +i\sin \theta \right)\\ 6-3i=3\sqrt{5}\left(\cos {333.43}^0+i\sin {333.43}^0\right)\\ z=\frac{(1+\sqrt{3}i)}{(6-3i)}=\frac{2\left(\cos {60}^0+i\sin {60}^0\right)}{3\sqrt{5}\left(\cos {333.43}^0+i\sin {333.43}^0\right)}\end{array}$

The trigonometric form is $\frac{(1+\sqrt{3}i)}{(6-3i)}=\frac{2\left(\cos {60}^0+i\sin {60}^0\right)}{3\sqrt{5}\left(\cos {333.43}^0+i\sin {333.43}^0\right)}$

Conclusion:

Thus,the trigonometric form is $\frac{(1+\sqrt{3}i)}{(6-3i)}=\frac{2\left(\cos {60}^0+i\sin {60}^0\right)}{3\sqrt{5}\left(\cos {333.43}^0+i\sin {333.43}^0\right)}$

(b)

To determine

The quotientusing the trigonometric forms.

Answer to Problem 62E

The quotientusing the trigonometric form is $z=0.018+0.298i$

Explanation of Solution

Given info:

  $\frac{(1+\sqrt{3}i)}{(6-3i)}$

Formula used:

Let $z_1=r_1\left(\cos {\theta }_1+i\sin {\theta }_1\right)\text{ and }{z}_2=r_2\left(\cos {\theta }_2+i\sin {\theta }_2\right)$ we have

  $\begin{array}{l}{z}_1{z}_2=r_1{r}_2\left(\cos \left({\theta }_1+{\theta }_2\right)+i\sin \left({\theta }_1+{\theta }_2\right)\right)\\ \frac{z_1}{z_2}=\frac{r_1}{r_2}\left(\cos \left({\theta }_1-{\theta }_2\right)+i\sin \left({\theta }_1-{\theta }_2\right)\right)\end{array}$

Calculation:

From part a we have $\frac{(1+\sqrt{3}i)}{(6-3i)}=\frac{2\left(\cos {60}^0+i\sin {60}^0\right)}{3\sqrt{5}\left(\cos {333.43}^0+i\sin {333.43}^0\right)}$

We have

  $\begin{array}{l}z=\frac{(1+\sqrt{3}i)}{(6-3i)}=\frac{2\left(\cos {60}^0+i\sin {60}^0\right)}{3\sqrt{5}\left(\cos {333.43}^0+i\sin {333.43}^0\right)}\\ z_1=r_1\left(\cos {\theta }_1+i\sin {\theta }_1\right)\text{=}2\left(\cos {60}^0+i\sin {60}^0\right)\\ z_2=r_2\left(\cos {\theta }_2+i\sin {\theta }_2\right)=3\sqrt{5}\left(\cos {333.43}^0+i\sin {333.43}^0\right)\\ z=\frac{z_1}{z_2}=\frac{r_1}{r_2}\left(\cos \left({\theta }_1-{\theta }_2\right)+i\sin \left({\theta }_1-{\theta }_2\right)\right)\\ z=\frac{2}{3\sqrt{5}}\left(\cos \left({60}^0-{333.43}^0\right)+i\sin \left({60}^0-{333.43}^0\right)\right)\\ z=0.018+0.298i\end{array}$

The quotientusing the trigonometric formis $z=0.018+0.298i$

Conclusion:

Thus,the quotientusing the trigonometric form is $z=0.018+0.298i$

(c)

To determine

The quotient using the standard forms, and check your result with that of part (b).

Answer to Problem 62E

The quotient using standard form and trigonometric form are same.

Explanation of Solution

Given info:

  $\frac{(1+\sqrt{3}i)}{(6-3i)}$

Formula used:

We have the results

  $i^2=-1$

Calculation:

Quotient from part b is $z=0.018+0.298i$

The quotient using the standard forms is given by

  $\begin{array}{l}z=\frac{(1+\sqrt{3}i)}{(6-3i)}\\ z=\frac{(1+\sqrt{3}i)}{(6-3i)}\times \frac{(6+3i)}{(6+3i)}=\frac{6+3i+6\sqrt{3}i+3\sqrt{3}{i}^2}{6^2+3^2}\\ z=\frac{6+3i+6\sqrt{3}i-3\sqrt{3}}{45}\\ z=0.018+0.298i\end{array}$

Quotientusing standard form and trigonometric form are same.

Conclusion:

Thus,the quotient using standard form and trigonometric form are same.