Chapter 6.6, Problem 80E

(a)

To determine

Thefourth roots of $i$ using formula for roots.

Answer to Problem 80E

Thefourth roots of $i$ are $z_0=\cos \frac{\pi }{8}+i\sin \frac{\pi }{8},z_1=\cos \frac{5\pi }{8}+i\sin \frac{5\pi }{8},z_2=\cos \frac{9\pi }{8}+i\sin \frac{9\pi }{8}\text{ and }{z}_3=\cos \frac{13\pi }{8}+i\sin \frac{13\pi }{8}$

Explanation of Solution

Given info:

  $\sqrt[4]{i}$

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

  $\begin{array}{l}a=0,b=1\\ r=\sqrt{a^2+b^2}=\sqrt{0^2+1^2}=1\\ \theta ={\tan }^{-1}\left(\frac{1}{0}\right)=\frac{\pi }{2}\\ i=\cos \frac{\pi }{2}+i\sin \frac{\pi }{2}\\ z_k=\sqrt[n]{r}\left(\cos \left(\frac{\theta +2\pi k}{n}\right)+i\sin \left(\frac{\theta +2\pi k}{n}\right)\right)\\ n=4,r=1,\theta =\frac{\pi }{2}\\ z_0=\sqrt[4]{1}\left(\cos \left(\frac{\frac{\pi }{2}+2\pi \times 0}{4}\right)+i\sin \left(\frac{\frac{\pi }{2}+2\pi \times 0}{4}\right)\right)\\ z_0=\cos \frac{\pi }{8}+i\sin \frac{\pi }{8}\\ z_1=\sqrt[4]{1}\left(\cos \left(\frac{\frac{\pi }{2}+2\pi \times 1}{4}\right)+i\sin \left(\frac{\frac{\pi }{2}+2\pi \times 1}{4}\right)\right)\\ z_1=\cos \frac{5\pi }{8}+i\sin \frac{5\pi }{8}\\ z_2=\sqrt[4]{1}\left(\cos \left(\frac{\frac{\pi }{2}+2\pi \times 2}{4}\right)+i\sin \left(\frac{\frac{\pi }{2}+2\pi \times 2}{4}\right)\right)\\ z_2=\cos \frac{9\pi }{8}+i\sin \frac{9\pi }{8}\\ z_3=\sqrt[4]{1}\left(\cos \left(\frac{\frac{\pi }{2}+2\pi \times 3}{4}\right)+i\sin \left(\frac{\frac{\pi }{2}+2\pi \times 3}{4}\right)\right)\\ z_3=\cos \frac{13\pi }{8}+i\sin \frac{13\pi }{8}\end{array}$

Thefourth roots of $i$ are $z_0=\cos \frac{\pi }{8}+i\sin \frac{\pi }{8},z_1=\cos \frac{5\pi }{8}+i\sin \frac{5\pi }{8},z_2=\cos \frac{9\pi }{8}+i\sin \frac{9\pi }{8}\text{ and }{z}_3=\cos \frac{13\pi }{8}+i\sin \frac{13\pi }{8}$

Conclusion:

Thus,the fourth roots of $i$ are $z_0=\cos \frac{\pi }{8}+i\sin \frac{\pi }{8},z_1=\cos \frac{5\pi }{8}+i\sin \frac{5\pi }{8},z_2=\cos \frac{9\pi }{8}+i\sin \frac{9\pi }{8}\text{ and }{z}_3=\cos \frac{13\pi }{8}+i\sin \frac{13\pi }{8}$

(b)

To determine

The standard form of each of the roots.

Answer to Problem 80E

The standard formof the roots are $z_0=0.92+0.38i,z_1=-0.38+0.92i,z_2=-0.92-0.38i\text{ and }{z}_3=0.38-0.92i$

Explanation of Solution

Given info:

Thefourth roots of $i$ are $z_0=\cos \frac{\pi }{8}+i\sin \frac{\pi }{8},z_1=\cos \frac{5\pi }{8}+i\sin \frac{5\pi }{8},z_2=\cos \frac{9\pi }{8}+i\sin \frac{9\pi }{8}\text{ and }{z}_3=\cos \frac{13\pi }{8}+i\sin \frac{13\pi }{8}$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 $z_0=\cos \frac{\pi }{8}+i\sin \frac{\pi }{8},z_1=\cos \frac{5\pi }{8}+i\sin \frac{5\pi }{8},z_2=\cos \frac{9\pi }{8}+i\sin \frac{9\pi }{8}\text{ and }{z}_3=\cos \frac{13\pi }{8}+i\sin \frac{13\pi }{8}$

We have

  $\begin{array}{l}{z}_0=\cos \frac{\pi }{8}+i\sin \frac{\pi }{8}=0.92+0.38i\\ z_1=\cos \frac{5\pi }{8}+i\sin \frac{5\pi }{8}=-0.38+0.92i\\ z_2=\cos \frac{9\pi }{8}+i\sin \frac{9\pi }{8}=-0.92-0.38i\\ z_3=\cos \frac{13\pi }{8}+i\sin \frac{13\pi }{8}=0.38-0.92i\end{array}$

The standard formof the roots are

  $z_0=0.92+0.38i,z_1=-0.38+0.92i,z_2=-0.92-0.38i\text{ and }{z}_3=0.38-0.92i$

Conclusion:

Thus,the standard formof the roots are $z_0=0.92+0.38i,z_1=-0.38+0.92i,z_2=-0.92-0.38i\text{ and }{z}_3=0.38-0.92i$ .

(c)

To determine

The roots graphically.

Answer to Problem 80E

Please refer calculation part.

Explanation of Solution

Given info:

Thefourth roots of $i$ are $z_0=0.92+0.38i,z_1=-0.38+0.92i,z_2=-0.92-0.38i\text{ and }{z}_3=0.38-0.92i$

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.