Chapter 6.6, Problem 75E

(a)

To determine

Thecube roots of $-\frac{125}{2}\left(1+\sqrt{3}i\right)$ using formula for roots.

Answer to Problem 75E

Thecube roots $-\frac{125}{2}\left(1+\sqrt{3}i\right)$ are $z_0=5\left(\cos \frac{4\pi }{9}+i\sin \frac{4\pi }{9}\right),z_1=5\left(\cos \frac{10\pi }{9}+i\sin \frac{10\pi }{9}\right)\text{ and }{z}_2=5\left(\cos \frac{16\pi }{9}+i\sin \frac{16\pi }{9}\right)$

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

  $\begin{array}{l}a=-\frac{125}{2},b=-\frac{125\sqrt{3}}{2}\\ r=\sqrt{a^2+b^2}=\sqrt{{\left(-\frac{125}{2}\right)}^2+{\left(-\frac{125\sqrt{3}}{2}\right)}^2}=125\\ \theta ={\tan }^{-1}\left(\frac{-\frac{125\sqrt{3}}{2}}{-\frac{125}{2}}\right)=\frac{4\pi }{3}\\ -\frac{125}{2}\left(1+\sqrt{3}i\right)=125\left(\cos \frac{4\pi }{3}+i\sin \frac{4\pi }{3}\right)\\ z_k=\sqrt[n]{r}\left(\cos \left(\frac{\theta +2\pi k}{n}\right)+i\sin \left(\frac{\theta +2\pi k}{n}\right)\right)\\ n=3,r=125,\theta =\frac{4\pi }{3}\\ z_0=\sqrt[3]{125}\left(\cos \left(\frac{\frac{4\pi }{3}+2\pi \times 0}{3}\right)+i\sin \left(\frac{\frac{4\pi }{3}+2\pi \times 0}{3}\right)\right)\\ z_0=5\left(\cos \frac{4\pi }{9}+i\sin \frac{4\pi }{9}\right)\\ z_1=\sqrt[3]{125}\left(\cos \left(\frac{\frac{4\pi }{3}+2\pi \times 1}{3}\right)+i\sin \left(\frac{\frac{4\pi }{3}+2\pi \times 1}{3}\right)\right)\\ z_1=5\left(\cos \frac{10\pi }{9}+i\sin \frac{10\pi }{9}\right)\\ z_2=\sqrt[3]{125}\left(\cos \left(\frac{\frac{4\pi }{3}+2\pi \times 2}{3}\right)+i\sin \left(\frac{\frac{4\pi }{3}+2\pi \times 2}{3}\right)\right)\\ z_2=5\left(\cos \frac{16\pi }{9}+i\sin \frac{16\pi }{9}\right)\end{array}$

Thecube roots of $-\frac{125}{2}\left(1+\sqrt{3}i\right)$ are $z_0=5\left(\cos \frac{4\pi }{9}+i\sin \frac{4\pi }{9}\right),z_1=5\left(\cos \frac{10\pi }{9}+i\sin \frac{10\pi }{9}\right)\text{ and }{z}_2=5\left(\cos \frac{16\pi }{9}+i\sin \frac{16\pi }{9}\right)$

Conclusion:

Thus,the cube roots of $-\frac{125}{2}\left(1+\sqrt{3}i\right)$ are $z_0=5\left(\cos \frac{4\pi }{9}+i\sin \frac{4\pi }{9}\right),z_1=5\left(\cos \frac{10\pi }{9}+i\sin \frac{10\pi }{9}\right)\text{ and }{z}_2=5\left(\cos \frac{16\pi }{9}+i\sin \frac{16\pi }{9}\right)$

(b)

To determine

The standard form of each of the roots.

Answer to Problem 75E

The standard formof the roots are $z_0=0.87+4.92i,z_1=-4.70-1.71i\text{ and }{z}_2=3.83-3.21i$

Explanation of Solution

Given info:

Thecube roots of $-\frac{125}{2}\left(1+\sqrt{3}i\right)$ are $z_0=5\left(\cos \frac{4\pi }{9}+i\sin \frac{4\pi }{9}\right),z_1=5\left(\cos \frac{10\pi }{9}+i\sin \frac{10\pi }{9}\right)\text{ and }{z}_2=5\left(\cos \frac{16\pi }{9}+i\sin \frac{16\pi }{9}\right)$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 $z_0=5\left(\cos \frac{4\pi }{9}+i\sin \frac{4\pi }{9}\right),z_1=5\left(\cos \frac{10\pi }{9}+i\sin \frac{10\pi }{9}\right)\text{ and }{z}_2=5\left(\cos \frac{16\pi }{9}+i\sin \frac{16\pi }{9}\right)$

We have

  $\begin{array}{l}{z}_0=5\left(\cos \frac{4\pi }{9}+i\sin \frac{4\pi }{9}\right)=0.87+4.92i\\ z_1=5\left(\cos \frac{10\pi }{9}+i\sin \frac{10\pi }{9}\right)=-4.70-1.71i\\ z_2=5\left(\cos \frac{16\pi }{9}+i\sin \frac{16\pi }{9}\right)=3.83-3.21i\end{array}$

The standard formof the roots are

  $z_0=0.87+4.92i,z_1=-4.70-1.71i\text{ and }{z}_2=3.83-3.21i$

Conclusion:

Thus,the standard formof the roots are $z_0=0.87+4.92i,z_1=-4.70-1.71i\text{ and }{z}_2=3.83-3.21i$ .

(c)

To determine

The roots graphically.

Answer to Problem 75E

Please refer calculation part.

Explanation of Solution

Given info:

Thecube roots of $-\frac{125}{2}\left(1+\sqrt{3}i\right)$ are $z_0=0.87+4.92i,z_1=-4.70-1.71i\text{ and }{z}_2=3.83-3.21i$

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.