Chapter 6.6, Problem 77E

(a)

To determine

Thesquare roots of $-25i$ using formula for roots.

Answer to Problem 77E

Thesquare roots of $-25i$ are $z_0=5\left(\cos \frac{3\pi }{4}+i\sin \frac{3\pi }{4}\right)\text{ and }{z}_1=5\left(\cos \frac{7\pi }{4}+i\sin \frac{7\pi }{4}\right)$

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

  $\begin{array}{l}a=0,b=-25\\ r=\sqrt{a^2+b^2}=\sqrt{0^2+{\left(-25\right)}^2}=25\\ \theta ={\tan }^{-1}\left(\frac{-25}{0}\right)=\frac{3\pi }{2}\\ -25i=25\left(\cos \frac{3\pi }{2}+i\sin \frac{3\pi }{2}\right)\\ z_k=\sqrt[n]{r}\left(\cos \left(\frac{\theta +2\pi k}{n}\right)+i\sin \left(\frac{\theta +2\pi k}{n}\right)\right)\\ n=2,r=25,\theta =\frac{3\pi }{2}\\ z_0=\sqrt[2]{25}\left(\cos \left(\frac{\frac{3\pi }{2}+2\pi \times 0}{2}\right)+i\sin \left(\frac{\frac{\pi }{2}+2\pi \times 0}{2}\right)\right)\\ z_0=5\left(\cos \frac{3\pi }{4}+i\sin \frac{3\pi }{4}\right)\\ z_1=\sqrt[2]{25}\left(\cos \left(\frac{\frac{3\pi }{2}+2\pi \times 1}{2}\right)+i\sin \left(\frac{\frac{\pi }{2}+2\pi \times 1}{2}\right)\right)\\ z_1=5\left(\cos \frac{7\pi }{4}+i\sin \frac{7\pi }{4}\right)\end{array}$

Thesquare roots of $-25i$ are $z_0=5\left(\cos \frac{3\pi }{4}+i\sin \frac{3\pi }{4}\right)\text{ and }{z}_1=5\left(\cos \frac{7\pi }{4}+i\sin \frac{7\pi }{4}\right)$

Conclusion:

Thus,the square roots of $-25i$ are $z_0=5\left(\cos \frac{3\pi }{4}+i\sin \frac{3\pi }{4}\right)\text{ and }{z}_1=5\left(\cos \frac{7\pi }{4}+i\sin \frac{7\pi }{4}\right)$

(b)

To determine

The standard form of each of the roots.

Answer to Problem 77E

The standard formof the roots are $z_0=-3.53+3.53i\text{ and }{z}_1=3.53-3.53i$

Explanation of Solution

Given info:

Thesquare roots of $-25i$ are $z_0=5\left(\cos \frac{3\pi }{4}+i\sin \frac{3\pi }{4}\right)\text{ and }{z}_1=5\left(\cos \frac{7\pi }{4}+i\sin \frac{7\pi }{4}\right)$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 $z_0=5\left(\cos \frac{3\pi }{4}+i\sin \frac{3\pi }{4}\right)\text{ and }{z}_1=5\left(\cos \frac{7\pi }{4}+i\sin \frac{7\pi }{4}\right)$

We have

  $\begin{array}{l}{z}_0=5\left(\cos \frac{3\pi }{4}+i\sin \frac{3\pi }{4}\right)=-3.53+3.53i\\ z_1=5\left(\cos \frac{7\pi }{4}+i\sin \frac{7\pi }{4}\right)=3.53-3.53i\end{array}$

The standard formof the roots are $z_0=-3.53+3.53i\text{ and }{z}_1=3.53-3.53i$

Conclusion:

Thus,the standard formof the roots are $z_0=-3.53+3.53i\text{ and }{z}_1=3.53-3.53i$ .

(c)

To determine

The roots graphically.

Answer to Problem 77E

Please refer calculation part.

Explanation of Solution

Given info:

Thesquare roots of $-25i$ are $z_0=-3.53+3.53i\text{ and }{z}_1=3.53-3.53i$

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.