Chapter 6, Problem 125RE

(a)

To determine

The sixth roots of $-729i$ using formula for roots.

Answer to Problem 125RE

The sixth roots of $-729i$ are $\begin{array}{l}{z}_0=3\left(\cos \frac{3\pi }{12}+i\sin \frac{3\pi }{12}\right),z_1=3\left(\cos \frac{7\pi }{12}+i\sin \frac{7\pi }{12}\right),z_2=3\left(\cos \frac{11\pi }{12}+i\sin \frac{11\pi }{12}\right),z_3=3\left(\cos \frac{15\pi }{12}+i\sin \frac{15\pi }{12}\right)\text{ ,}\\ z_4=3\left(\cos \frac{19\pi }{12}+i\sin \frac{19\pi }{12}\right)\text{and }{z}_5=3\left(\cos \frac{23\pi }{12}+i\sin \frac{23\pi }{12}\right)\end{array}$

Explanation of Solution

Given info:

  $\sqrt[6]{-729i}$

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 $\theta ={\tan }^{-1}\left(\frac{b}{a}\right)$

For a positive integer n the complex number $z=r\left(\cos \theta +i\sin \theta \right)$ has exactly n distinct nth roots given by

  $z_k=\sqrt[n]{r}\left(\cos \left(\frac{\theta +2\pi k}{n}\right)+i\sin \left(\frac{\theta +2\pi k}{n}\right)\right)$

where k = 0, 1, 2, . . . ,,n -1.

Calculation:

We have

  $\begin{array}{l}a=0,b=-729\\ r=\sqrt{a^2+b^2}=\sqrt{0^2+{\left(-729\right)}^2}=729\\ \theta ={\tan }^{-1}\left(\frac{-729}{0}\right)=\frac{3\pi }{2}\end{array}$

  $-729i=729\left(\cos \frac{3\pi }{2}+i\sin \frac{3\pi }{2}\right)$

  $\begin{array}{l}{z}_k=\sqrt[n]{r}\left(\cos \left(\frac{\theta +2\pi k}{n}\right)+i\sin \left(\frac{\theta +2\pi k}{n}\right)\right)\\ n=6,r=729,\theta =\frac{3\pi }{2}\end{array}$

  $\begin{array}{l}{z}_0=\sqrt[6]{729}\left(\cos \left(\frac{\frac{3\pi }{2}+2\pi \times 0}{6}\right)+i\sin \left(\frac{\frac{3\pi }{2}+2\pi \times 0}{6}\right)\right)\\ z_0=3\left(\cos \frac{3\pi }{12}+i\sin \frac{3\pi }{12}\right)\end{array}$

  $\begin{array}{l}{z}_1=\sqrt[6]{729}\left(\cos \left(\frac{\frac{3\pi }{2}+2\pi \times 1}{6}\right)+i\sin \left(\frac{\frac{3\pi }{2}+2\pi \times 1}{6}\right)\right)\\ z_1=3\left(\cos \frac{7\pi }{12}+i\sin \frac{7\pi }{12}\right)\end{array}$

  $\begin{array}{l}{z}_2=\sqrt[6]{729}\left(\cos \left(\frac{\frac{3\pi }{2}+2\pi \times 2}{6}\right)+i\sin \left(\frac{\frac{3\pi }{2}+2\pi \times 2}{6}\right)\right)\\ z_2=3\left(\cos \frac{11\pi }{12}+i\sin \frac{11\pi }{12}\right)\end{array}$

  $\begin{array}{l}{z}_3=\sqrt[6]{729}\left(\cos \left(\frac{\frac{3\pi }{2}+2\pi \times 3}{6}\right)+i\sin \left(\frac{\frac{3\pi }{2}+2\pi \times 3}{6}\right)\right)\\ z_3=3\left(\cos \frac{15\pi }{12}+i\sin \frac{15\pi }{12}\right)\end{array}$

  $\begin{array}{l}{z}_4=\sqrt[6]{729}\left(\cos \left(\frac{\frac{3\pi }{2}+2\pi \times 4}{6}\right)+i\sin \left(\frac{\frac{3\pi }{2}+2\pi \times 4}{6}\right)\right)\\ z_4=3\left(\cos \frac{19\pi }{12}+i\sin \frac{19\pi }{12}\right)\end{array}$

  $\begin{array}{l}{z}_5=\sqrt[6]{729}\left(\cos \left(\frac{\frac{3\pi }{2}+2\pi \times 5}{6}\right)+i\sin \left(\frac{\frac{3\pi }{2}+2\pi \times 5}{6}\right)\right)\\ z_5=3\left(\cos \frac{23\pi }{12}+i\sin \frac{23\pi }{12}\right)\end{array}$

The sixth roots of $-729i$ are $\begin{array}{l}{z}_0=3\left(\cos \frac{3\pi }{12}+i\sin \frac{3\pi }{12}\right),z_1=3\left(\cos \frac{7\pi }{12}+i\sin \frac{7\pi }{12}\right),z_2=3\left(\cos \frac{11\pi }{12}+i\sin \frac{11\pi }{12}\right),z_3=3\left(\cos \frac{15\pi }{12}+i\sin \frac{15\pi }{12}\right)\text{ ,}\\ z_4=3\left(\cos \frac{19\pi }{12}+i\sin \frac{19\pi }{12}\right)\text{and }{z}_5=3\left(\cos \frac{23\pi }{12}+i\sin \frac{23\pi }{12}\right)\end{array}$

Conclusion:

Thus,the sixth roots of $-729i$ are $\begin{array}{l}{z}_0=3\left(\cos \frac{3\pi }{12}+i\sin \frac{3\pi }{12}\right),z_1=3\left(\cos \frac{7\pi }{12}+i\sin \frac{7\pi }{12}\right),z_2=3\left(\cos \frac{11\pi }{12}+i\sin \frac{11\pi }{12}\right),z_3=3\left(\cos \frac{15\pi }{12}+i\sin \frac{15\pi }{12}\right)\text{ ,}\\ z_4=3\left(\cos \frac{19\pi }{12}+i\sin \frac{19\pi }{12}\right)\text{and }{z}_5=3\left(\cos \frac{23\pi }{12}+i\sin \frac{23\pi }{12}\right)\end{array}$

(b)

To determine

The standard form of each of the roots.

Answer to Problem 125RE

The standard formof the roots are $\begin{array}{l}{z}_0=2.12+2.12i,z_1=-0.78+2.90i,z_2=-2.90+0.78i,z_3=-2.12-2.12i,z_4=0.78-2.90i\\ \text{ and }{z}_5=2.90-0.78i\end{array}$

Explanation of Solution

Given info:

The sixth roots of $-729i$ are $\begin{array}{l}{z}_0=3\left(\cos \frac{3\pi }{12}+i\sin \frac{3\pi }{12}\right),z_1=3\left(\cos \frac{7\pi }{12}+i\sin \frac{7\pi }{12}\right),z_2=3\left(\cos \frac{11\pi }{12}+i\sin \frac{11\pi }{12}\right),z_3=3\left(\cos \frac{15\pi }{12}+i\sin \frac{15\pi }{12}\right)\text{ ,}\\ z_4=3\left(\cos \frac{19\pi }{12}+i\sin \frac{19\pi }{12}\right)\text{and }{z}_5=3\left(\cos \frac{23\pi }{12}+i\sin \frac{23\pi }{12}\right)\end{array}$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:

The roots from part a are $\begin{array}{l}{z}_0=3\left(\cos \frac{3\pi }{12}+i\sin \frac{3\pi }{12}\right),z_1=3\left(\cos \frac{7\pi }{12}+i\sin \frac{7\pi }{12}\right),z_2=3\left(\cos \frac{11\pi }{12}+i\sin \frac{11\pi }{12}\right),z_3=3\left(\cos \frac{15\pi }{12}+i\sin \frac{15\pi }{12}\right)\text{ ,}\\ z_4=3\left(\cos \frac{19\pi }{12}+i\sin \frac{19\pi }{12}\right)\text{and }{z}_5=3\left(\cos \frac{23\pi }{12}+i\sin \frac{23\pi }{12}\right)\end{array}$

We have

  $\begin{array}{l}{z}_0=3\left(\cos \frac{3\pi }{12}+i\sin \frac{3\pi }{12}\right)=2.12+2.12i\\ z_1=3\left(\cos \frac{7\pi }{12}+i\sin \frac{7\pi }{12}\right)=-0.78+2.90i\\ z_2=3\left(\cos \frac{11\pi }{12}+i\sin \frac{11\pi }{12}\right)=-2.90+0.78i\\ z_3=3\left(\cos \frac{15\pi }{12}+i\sin \frac{15\pi }{12}\right)=-2.12-2.12i\\ z_4=3\left(\cos \frac{19\pi }{12}+i\sin \frac{19\pi }{12}\right)=0.78-2.90i\\ z_5=3\left(\cos \frac{23\pi }{12}+i\sin \frac{23\pi }{12}\right)=2.90-0.78i\end{array}$

The standard formof the roots are

  $\begin{array}{l}{z}_0=2.12+2.12i,z_1=-0.78+2.90i,z_2=-2.90+0.78i,z_3=-2.12-2.12i,z_4=0.78-2.90i\\ \text{ and }{z}_5=2.90-0.78i\end{array}$

Conclusion:

Thus,the standard formof the roots are $\begin{array}{l}{z}_0=2.12+2.12i,z_1=-0.78+2.90i,z_2=-2.90+0.78i,z_3=-2.12-2.12i,z_4=0.78-2.90i\\ \text{ and }{z}_5=2.90-0.78i\end{array}$

(c)

To determine

The roots graphically.

Answer to Problem 125RE

Please refer calculation part.

Explanation of Solution

Given info:

The sixth roots of $-729i$ are $\begin{array}{l}{z}_0=2.12+2.12i,z_1=-0.78+2.90i,z_2=-2.90+0.78i,z_3=-2.12-2.12i,z_4=0.78-2.90i\\ \text{ and }{z}_5=2.90-0.78i\end{array}$

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:

Graphical representation of roots are given by

  

Conclusion:

Thus,the graphical representation of roots are given.