Chapter 6.6, Problem 84E

(a)

To determine

Thefourth roots of $-4$ using formula for roots.

Answer to Problem 84E

Thefourth roots of $-4$ are $\begin{array}{l}{z}_0=\sqrt{2}\left(\cos \frac{\pi }{4}+i\sin \frac{\pi }{4}\right),z_1=\sqrt{2}\left(\cos \frac{3\pi }{4}+i\sin \frac{3\pi }{4}\right),z_2=\sqrt{2}\left(\cos \frac{5\pi }{4}+i\sin \frac{5\pi }{4}\right)\\ \text{ and }{z}_3=\sqrt{2}\left(\cos \frac{7\pi }{4}+i\sin \frac{7\pi }{4}\right)\end{array}$

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

  $\begin{array}{l}a=-4,b=0\\ r=\sqrt{a^2+b^2}=\sqrt{{\left(-4\right)}^2+0^2}=4\\ \theta ={\tan }^{-1}\left(\frac{0}{-4}\right)=\pi \\ -4=4\left(\cos \pi +i\sin \pi \right)\\ z_k=\sqrt[n]{r}\left(\cos \left(\frac{\theta +2\pi k}{n}\right)+i\sin \left(\frac{\theta +2\pi k}{n}\right)\right)\\ n=4,r=4,\theta =\pi \\ z_0=\sqrt[4]{4}\left(\cos \left(\frac{\pi +2\pi \times 0}{4}\right)+i\sin \left(\frac{\pi +2\pi \times 0}{4}\right)\right)\\ z_0=\sqrt{2}\left(\cos \frac{\pi }{4}+i\sin \frac{\pi }{4}\right)\\ z_1=\sqrt[4]{4}\left(\cos \left(\frac{\pi +2\pi \times 1}{4}\right)+i\sin \left(\frac{\pi +2\pi \times 1}{4}\right)\right)\\ z_1=\sqrt{2}\left(\cos \frac{3\pi }{4}+i\sin \frac{3\pi }{4}\right)\\ z_2=\sqrt[4]{4}\left(\cos \left(\frac{\pi +2\pi \times 2}{4}\right)+i\sin \left(\frac{\pi +2\pi \times 2}{4}\right)\right)\\ z_2=\sqrt{2}\left(\cos \frac{5\pi }{4}+i\sin \frac{5\pi }{4}\right)\\ z_3=\sqrt[4]{4}\left(\cos \left(\frac{\pi +2\pi \times 3}{4}\right)+i\sin \left(\frac{\pi +2\pi \times 3}{4}\right)\right)\\ z_3=\sqrt{2}\left(\cos \frac{7\pi }{4}+i\sin \frac{7\pi }{4}\right)\end{array}$

Thefourth roots of $-4$ are $\begin{array}{l}{z}_0=\sqrt{2}\left(\cos \frac{\pi }{4}+i\sin \frac{\pi }{4}\right),z_1=\sqrt{2}\left(\cos \frac{3\pi }{4}+i\sin \frac{3\pi }{4}\right),z_2=\sqrt{2}\left(\cos \frac{5\pi }{4}+i\sin \frac{5\pi }{4}\right)\\ \text{ and }{z}_3=\sqrt{2}\left(\cos \frac{7\pi }{4}+i\sin \frac{7\pi }{4}\right)\end{array}$

Conclusion:

Thus,the fourth roots of $-4$ are $\begin{array}{l}{z}_0=\sqrt{2}\left(\cos \frac{\pi }{4}+i\sin \frac{\pi }{4}\right),z_1=\sqrt{2}\left(\cos \frac{3\pi }{4}+i\sin \frac{3\pi }{4}\right),z_2=\sqrt{2}\left(\cos \frac{5\pi }{4}+i\sin \frac{5\pi }{4}\right)\\ \text{ and }{z}_3=\sqrt{2}\left(\cos \frac{7\pi }{4}+i\sin \frac{7\pi }{4}\right)\end{array}$

(b)

To determine

The standard form of each of the roots.

Answer to Problem 84E

The standard formof the roots are $z_0=1+i,z_1=-1+i,z_2=-1-i\text{ and }{z}_3=1-i$

Explanation of Solution

Given info:

Thefourth roots of $-4$ are $\begin{array}{l}{z}_0=\sqrt{2}\left(\cos \frac{\pi }{4}+i\sin \frac{\pi }{4}\right),z_1=\sqrt{2}\left(\cos \frac{3\pi }{4}+i\sin \frac{3\pi }{4}\right),z_2=\sqrt{2}\left(\cos \frac{5\pi }{4}+i\sin \frac{5\pi }{4}\right)\\ \text{ and }{z}_3=\sqrt{2}\left(\cos \frac{7\pi }{4}+i\sin \frac{7\pi }{4}\right)\end{array}$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 $\begin{array}{l}{z}_0=\sqrt{2}\left(\cos \frac{\pi }{4}+i\sin \frac{\pi }{4}\right),z_1=\sqrt{2}\left(\cos \frac{3\pi }{4}+i\sin \frac{3\pi }{4}\right),z_2=\sqrt{2}\left(\cos \frac{5\pi }{4}+i\sin \frac{5\pi }{4}\right)\\ \text{ and }{z}_3=\sqrt{2}\left(\cos \frac{7\pi }{4}+i\sin \frac{7\pi }{4}\right)\end{array}$

We have

  $\begin{array}{l}{z}_0=\sqrt{2}\left(\cos \frac{\pi }{4}+i\sin \frac{\pi }{4}\right)=1+i\\ z_1=\sqrt{2}\left(\cos \frac{3\pi }{4}+i\sin \frac{3\pi }{4}\right)=-1+i\\ z_2=\sqrt{2}\left(\cos \frac{5\pi }{4}+i\sin \frac{5\pi }{4}\right)=-1-i\\ z_3=\sqrt{2}\left(\cos \frac{7\pi }{4}+i\sin \frac{7\pi }{4}\right)=1-i\end{array}$

The standard formof the roots are

  $z_0=1+i,z_1=-1+i,z_2=-1-i\text{ and }{z}_3=1-i$

Conclusion:

Thus,the standard formof the roots are $z_0=1+i,z_1=-1+i,z_2=-1-i\text{ and }{z}_3=1-i$ .

(c)

To determine

The roots graphically.

Answer to Problem 84E

Please refer calculation part.

Explanation of Solution

Given info:

Thefourth roots of $-4$ are $z_0=1+i,z_1=-1+i,z_2=-1-i\text{ and }{z}_3=1-i$

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

Calculation:

Graphical representation of roots are given by

  

Conclusion:

Thus,the graphical representation of roots are given.