Chapter 6.6, Problem 86E

(a)

To determine

Thesixth roots of $64i$ using formula for roots.

Answer to Problem 86E

Thesixth roots of $64i$ are $\begin{array}{l}{z}_0=2\left(\cos \frac{\pi }{12}+i\sin \frac{\pi }{12}\right),z_1=2\left(\cos \frac{5\pi }{12}+i\sin \frac{5\pi }{12}\right),z_2=2\left(\cos \frac{9\pi }{12}+i\sin \frac{9\pi }{12}\right),z_3=2\left(\cos \frac{13\pi }{12}+i\sin \frac{13\pi }{12}\right)\text{ ,}\\ z_4=2\left(\cos \frac{17\pi }{12}+i\sin \frac{17\pi }{12}\right)\text{and }{z}_5=2\left(\cos \frac{21\pi }{12}+i\sin \frac{21\pi }{12}\right)\end{array}$

Explanation of Solution

Given info:

  $\sqrt[6]{64i}$

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

  $\begin{array}{l}a=0,b=64\\ r=\sqrt{a^2+b^2}=\sqrt{0^2+{64}^2}=64\end{array}$

  $\begin{array}{l}\theta ={\tan }^{-1}\left(\frac{64}{0}\right)=\frac{\pi }{2}\\ 64i=64\left(\cos \frac{\pi }{2}+i\sin \frac{\pi }{2}\right)\end{array}$

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

  $n=6,r=64,\theta =\frac{\pi }{2}$

  $z_0=\sqrt[6]{64}\left(\cos \left(\frac{\frac{\pi }{2}+2\pi \times 0}{6}\right)+i\sin \left(\frac{\frac{\pi }{2}+2\pi \times 0}{6}\right)\right)$

  $z_0=2\left(\cos \frac{\pi }{12}+i\sin \frac{\pi }{12}\right)$

  $z_1=\sqrt[6]{64}\left(\cos \left(\frac{\frac{\pi }{2}+2\pi \times 1}{6}\right)+i\sin \left(\frac{\frac{\pi }{2}+2\pi \times 1}{6}\right)\right)$

  $z_1=2\left(\cos \frac{5\pi }{12}+i\sin \frac{5\pi }{12}\right)$

  $z_2=\sqrt[6]{64}\left(\cos \left(\frac{\frac{\pi }{2}+2\pi \times 2}{6}\right)+i\sin \left(\frac{\frac{\pi }{2}+2\pi \times 2}{6}\right)\right)$

  $z_2=2\left(\cos \frac{9\pi }{12}+i\sin \frac{9\pi }{12}\right)$

  $z_3=\sqrt[6]{64}\left(\cos \left(\frac{\frac{\pi }{2}+2\pi \times 3}{6}\right)+i\sin \left(\frac{\frac{\pi }{2}+2\pi \times 3}{6}\right)\right)$

  $z_3=2\left(\cos \frac{13\pi }{12}+i\sin \frac{13\pi }{12}\right)$

  $z_4=\sqrt[6]{64}\left(\cos \left(\frac{\frac{\pi }{2}+2\pi \times 4}{6}\right)+i\sin \left(\frac{\frac{\pi }{2}+2\pi \times 4}{6}\right)\right)$

  $z_4=2\left(\cos \frac{17\pi }{12}+i\sin \frac{17\pi }{12}\right)$

  $z_5=\sqrt[6]{64}\left(\cos \left(\frac{\frac{\pi }{2}+2\pi \times 5}{6}\right)+i\sin \left(\frac{\frac{\pi }{2}+2\pi \times 5}{6}\right)\right)$

  $z_5=2\left(\cos \frac{21\pi }{12}+i\sin \frac{21\pi }{12}\right)$

Thesixth roots of $64i$ are $\begin{array}{l}{z}_0=2\left(\cos \frac{\pi }{12}+i\sin \frac{\pi }{12}\right),z_1=2\left(\cos \frac{5\pi }{12}+i\sin \frac{5\pi }{12}\right),z_2=2\left(\cos \frac{9\pi }{12}+i\sin \frac{9\pi }{12}\right),z_3=2\left(\cos \frac{13\pi }{12}+i\sin \frac{13\pi }{12}\right)\text{ ,}\\ z_4=2\left(\cos \frac{17\pi }{12}+i\sin \frac{17\pi }{12}\right)\text{and }{z}_5=2\left(\cos \frac{21\pi }{12}+i\sin \frac{21\pi }{12}\right)\end{array}$

Conclusion:

Thus,the sixth roots of $64i$ are $\begin{array}{l}{z}_0=2\left(\cos \frac{\pi }{12}+i\sin \frac{\pi }{12}\right),z_1=2\left(\cos \frac{5\pi }{12}+i\sin \frac{5\pi }{12}\right),z_2=2\left(\cos \frac{9\pi }{12}+i\sin \frac{9\pi }{12}\right),z_3=2\left(\cos \frac{13\pi }{12}+i\sin \frac{13\pi }{12}\right)\text{ ,}\\ z_4=2\left(\cos \frac{17\pi }{12}+i\sin \frac{17\pi }{12}\right)\text{and }{z}_5=2\left(\cos \frac{21\pi }{12}+i\sin \frac{21\pi }{12}\right)\end{array}$

(b)

To determine

The standard form of each of the roots.

Answer to Problem 86E

The standard formof the roots are $\begin{array}{l}{z}_0=1.93+0.52i,z_1=0.52+1.93i,z_2=-1.41+1.41i,z_3=-1.93-0.52i,z_4=-0.52-1.93i\\ \text{ and }{z}_5=1.41-1.41i\end{array}$

Explanation of Solution

Given info:

Thesixth roots of $64i$ are $\begin{array}{l}{z}_0=2\left(\cos \frac{\pi }{12}+i\sin \frac{\pi }{12}\right),z_1=2\left(\cos \frac{5\pi }{12}+i\sin \frac{5\pi }{12}\right),z_2=2\left(\cos \frac{9\pi }{12}+i\sin \frac{9\pi }{12}\right),z_3=2\left(\cos \frac{13\pi }{12}+i\sin \frac{13\pi }{12}\right)\text{ ,}\\ z_4=2\left(\cos \frac{17\pi }{12}+i\sin \frac{17\pi }{12}\right)\text{and }{z}_5=2\left(\cos \frac{21\pi }{12}+i\sin \frac{21\pi }{12}\right)\end{array}$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 $\begin{array}{l}{z}_0=2\left(\cos \frac{\pi }{12}+i\sin \frac{\pi }{12}\right),z_1=2\left(\cos \frac{5\pi }{12}+i\sin \frac{5\pi }{12}\right),z_2=2\left(\cos \frac{9\pi }{12}+i\sin \frac{9\pi }{12}\right),z_3=2\left(\cos \frac{13\pi }{12}+i\sin \frac{13\pi }{12}\right)\text{ ,}\\ z_4=2\left(\cos \frac{17\pi }{12}+i\sin \frac{17\pi }{12}\right)\text{and }{z}_5=2\left(\cos \frac{21\pi }{12}+i\sin \frac{21\pi }{12}\right)\end{array}$

We have

  $\begin{array}{l}{z}_0=2\left(\cos \frac{\pi }{12}+i\sin \frac{\pi }{12}\right)=1.93+0.52i\\ z_1=2\left(\cos \frac{5\pi }{12}+i\sin \frac{5\pi }{12}\right)=0.52+1.93i\\ z_2=2\left(\cos \frac{9\pi }{12}+i\sin \frac{9\pi }{12}\right)=-1.41+1.41i\\ z_3=2\left(\cos \frac{13\pi }{12}+i\sin \frac{13\pi }{12}\right)=-1.93-0.52i\\ z_4=2\left(\cos \frac{17\pi }{12}+i\sin \frac{17\pi }{12}\right)=-0.52-1.93i\\ z_5=2\left(\cos \frac{21\pi }{12}+i\sin \frac{21\pi }{12}\right)=1.41-1.41i\end{array}$

The standard formof the roots are

  $\begin{array}{l}{z}_0=1.93+0.52i,z_1=0.52+1.93i,z_2=-1.41+1.41i,z_3=-1.93-0.52i,z_4=-0.52-1.93i\\ \text{ and }{z}_5=1.41-1.41i\end{array}$

Conclusion:

Thus,the standard formof the roots are $\begin{array}{l}{z}_0=1.93+0.52i,z_1=0.52+1.93i,z_2=-1.41+1.41i,z_3=-1.93-0.52i,z_4=-0.52-1.93i\\ \text{ and }{z}_5=1.41-1.41i\end{array}$ .

(c)

To determine

The roots graphically.

Answer to Problem 86E

Please refer calculation part.

Explanation of Solution

Given info:

Thesixth roots of $64i$ are $\begin{array}{l}{z}_0=1.93+0.52i,z_1=0.52+1.93i,z_2=-1.41+1.41i,z_3=-1.93-0.52i,z_4=-0.52-1.93i\\ \text{ and }{z}_5=1.41-1.41i\end{array}$

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.