Chapter 6.6, Problem 78E

(a)

To determine

Thefourth roots of $625i$ using formula for roots.

Answer to Problem 78E

Thefourth roots of $625i$ are $z_0=5\left(\cos \frac{\pi }{8}+i\sin \frac{\pi }{8}\right),z_1=5\left(\cos \frac{5\pi }{8}+i\sin \frac{5\pi }{8}\right),z_2=5\left(\cos \frac{9\pi }{8}+i\sin \frac{9\pi }{8}\right)\text{ and }{z}_3=5\left(\cos \frac{13\pi }{8}+i\sin \frac{13\pi }{8}\right)$

Explanation of Solution

Given info:

  $\sqrt[4]{625i}$

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

  $\begin{array}{l}a=0,b=625\\ r=\sqrt{a^2+b^2}=\sqrt{0^2+{625}^2}=625\\ \theta ={\tan }^{-1}\left(\frac{625}{0}\right)=\frac{\pi }{2}\\ 625i=625\left(\cos \frac{\pi }{2}+i\sin \frac{\pi }{2}\right)\\ z_k=\sqrt[n]{r}\left(\cos \left(\frac{\theta +2\pi k}{n}\right)+i\sin \left(\frac{\theta +2\pi k}{n}\right)\right)\\ n=4,r=625,\theta =\frac{\pi }{2}\\ z_0=\sqrt[4]{625}\left(\cos \left(\frac{\frac{\pi }{2}+2\pi \times 0}{4}\right)+i\sin \left(\frac{\frac{\pi }{2}+2\pi \times 0}{4}\right)\right)\\ z_0=5\left(\cos \frac{\pi }{8}+i\sin \frac{\pi }{8}\right)\\ z_1=\sqrt[4]{625}\left(\cos \left(\frac{\frac{\pi }{2}+2\pi \times 1}{4}\right)+i\sin \left(\frac{\frac{\pi }{2}+2\pi \times 1}{4}\right)\right)\\ z_1=5\left(\cos \frac{5\pi }{8}+i\sin \frac{5\pi }{8}\right)\\ z_2=\sqrt[4]{625}\left(\cos \left(\frac{\frac{\pi }{2}+2\pi \times 2}{4}\right)+i\sin \left(\frac{\frac{\pi }{2}+2\pi \times 2}{4}\right)\right)\\ z_2=5\left(\cos \frac{9\pi }{8}+i\sin \frac{9\pi }{8}\right)\\ z_3=\sqrt[4]{625}\left(\cos \left(\frac{\frac{\pi }{2}+2\pi \times 3}{4}\right)+i\sin \left(\frac{\frac{\pi }{2}+2\pi \times 3}{4}\right)\right)\\ z_3=5\left(\cos \frac{13\pi }{8}+i\sin \frac{13\pi }{8}\right)\end{array}$

Thefourth roots of $625i$ are $z_0=5\left(\cos \frac{\pi }{8}+i\sin \frac{\pi }{8}\right),z_1=5\left(\cos \frac{5\pi }{8}+i\sin \frac{5\pi }{8}\right),z_2=5\left(\cos \frac{9\pi }{8}+i\sin \frac{9\pi }{8}\right)\text{ and }{z}_3=5\left(\cos \frac{13\pi }{8}+i\sin \frac{13\pi }{8}\right)$

Conclusion:

Thus,the fourth roots of $625i$ are $z_0=5\left(\cos \frac{\pi }{8}+i\sin \frac{\pi }{8}\right),z_1=5\left(\cos \frac{5\pi }{8}+i\sin \frac{5\pi }{8}\right),z_2=5\left(\cos \frac{9\pi }{8}+i\sin \frac{9\pi }{8}\right)\text{ and }{z}_3=5\left(\cos \frac{13\pi }{8}+i\sin \frac{13\pi }{8}\right)$

(b)

To determine

The standard form of each of the roots.

Answer to Problem 78E

The standard formof the roots are $z_0=4.62+1.91i,z_1=-1.91+4.62i,z_2=-4.62-1.91i\text{ and }{z}_3=1.91-4.62i$

Explanation of Solution

Given info:

Thefourth roots of $625i$ are $z_0=5\left(\cos \frac{\pi }{8}+i\sin \frac{\pi }{8}\right),z_1=5\left(\cos \frac{5\pi }{8}+i\sin \frac{5\pi }{8}\right),z_2=5\left(\cos \frac{9\pi }{8}+i\sin \frac{9\pi }{8}\right)\text{ and }{z}_3=5\left(\cos \frac{13\pi }{8}+i\sin \frac{13\pi }{8}\right)$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 $z_0=5\left(\cos \frac{\pi }{8}+i\sin \frac{\pi }{8}\right),z_1=5\left(\cos \frac{5\pi }{8}+i\sin \frac{5\pi }{8}\right),z_2=5\left(\cos \frac{9\pi }{8}+i\sin \frac{9\pi }{8}\right)\text{ and }{z}_3=5\left(\cos \frac{13\pi }{8}+i\sin \frac{13\pi }{8}\right)$

We have

  $\begin{array}{l}{z}_0=5\left(\cos \frac{\pi }{8}+i\sin \frac{\pi }{8}\right)=4.62+1.91i\\ z_1=5\left(\cos \frac{5\pi }{8}+i\sin \frac{5\pi }{8}\right)=-1.91+4.62i\\ z_2=5\left(\cos \frac{9\pi }{8}+i\sin \frac{9\pi }{8}\right)=-4.62-1.91i\\ z_3=5\left(\cos \frac{13\pi }{8}+i\sin \frac{13\pi }{8}\right)=1.91-4.62i\end{array}$

The standard formof the roots are

  $z_0=4.62+1.91i,z_1=-1.91+4.62i,z_2=-4.62-1.91i\text{ and }{z}_3=1.91-4.62i$

Conclusion:

Thus,the standard formof the roots are $z_0=4.62+1.91i,z_1=-1.91+4.62i,z_2=-4.62-1.91i\text{ and }{z}_3=1.91-4.62i$ .

(c)

To determine

The roots graphically.

Answer to Problem 78E

Please refer calculation part.

Explanation of Solution

Given info:

Thefourth roots of $625i$ are $z_0=4.62+1.91i,z_1=-1.91+4.62i,z_2=-4.62-1.91i\text{ and }{z}_3=1.91-4.62i$

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.