Chapter 6.6, Problem 111E

a)

To determine

To find: The cube roots of $-25i$

Answer to Problem 111E

The cube roots of $-25i$ are $5\left(cos\frac{\pi }{2}+i\sin \frac{\pi }{2}\right),5\left(\cos \frac{7\pi }{6}+i\sin \frac{7\pi }{6}\right)\&5\left(\cos \frac{11\pi }{6}+i\sin \frac{11\pi }{6}\right)$

Explanation of Solution

Given information: The complex number $-25i$

Formula used: $n^{th}$ roots of the complex number: $\sqrt[n]{r}\left(\cos \frac{\frac{3\pi }{2}+2\pi k}{6}+i\sin \frac{\frac{3\pi }{2}+2\pi k}{6}\right);k=0,1,2,3,\mathrm{...},n-1$

Calculation: Apply the $n^{th}$ roots formula to obtain the roots of the given number.

  $-25i=25\left(\cos \frac{3\pi }{2}+i\sin \frac{3\pi }{2}\right)$

  $\begin{array}{l}\sqrt[3]{25}\left(\cos \frac{\frac{3\pi }{2}+2\pi k}{3}+i\sin \frac{\frac{3\pi }{2}+2\pi k}{3}\right)=5\left(\cos \frac{3\pi +4\pi k}{6}+i\sin \frac{3\pi +4\pi k}{6}\right)\\ \therefore 5\left(\cos \frac{3\pi +4\pi \left(0\right)}{6}+i\sin \frac{3\pi +4\pi \left(0\right)}{6}\right)=5\left(\cos \frac{\pi }{2}+i\sin \frac{\pi }{2}\right)\\ \therefore 5\left(\cos \frac{3\pi +4\pi \left(1\right)}{6}+i\sin \frac{3\pi +4\pi \left(1\right)}{6}\right)=5\left(\cos \frac{7\pi }{6}+i\sin \frac{7\pi }{6}\right)\\ \therefore 5\left(\cos \frac{3\pi +4\pi \left(2\right)}{6}+i\sin \frac{3\pi +4\pi \left(2\right)}{6}\right)=5\left(\cos \frac{11\pi }{6}+i\sin \frac{11\pi }{6}\right)\end{array}$

b)

To determine

To find: The cube roots of $-25i$

Answer to Problem 111E

The roots of $-25i$ are in standard form: $2.9240i,-2.5323-1.4620i\&2.5323-1.4620i$

Explanation of Solution

Given information: The complex number $-25i$

Formula used: $n^{th}$ roots of the complex number: $\sqrt[n]{r}\left(\cos \frac{\frac{3\pi }{2}+2\pi k}{6}+i\sin \frac{\frac{3\pi }{2}+2\pi k}{6}\right);k=0,1,2,3,\mathrm{...},n-1$

Calculation: Apply the $n^{th}$ roots formula to obtain the roots of the given number.

  $-25i=25\left(\cos \frac{3\pi }{2}+i\sin \frac{3\pi }{2}\right)$

  $\begin{array}{l}\sqrt[3]{25}\left(\cos \frac{\frac{3\pi }{2}+2\pi k}{3}+i\sin \frac{\frac{3\pi }{2}+2\pi k}{3}\right)=5\left(\cos \frac{3\pi +4\pi k}{6}+i\sin \frac{3\pi +4\pi k}{6}\right)\\ \therefore 5\left(\cos \frac{3\pi +4\pi \left(0\right)}{6}+i\sin \frac{3\pi +4\pi \left(0\right)}{6}\right)=5\left(\cos \frac{\pi }{2}+i\sin \frac{\pi }{2}\right)\\ \therefore 5\left(\cos \frac{3\pi +4\pi \left(1\right)}{6}+i\sin \frac{3\pi +4\pi \left(1\right)}{6}\right)=5\left(\cos \frac{7\pi }{6}+i\sin \frac{7\pi }{6}\right)\\ \therefore 5\left(\cos \frac{3\pi +4\pi \left(2\right)}{6}+i\sin \frac{3\pi +4\pi \left(2\right)}{6}\right)=5\left(\cos \frac{11\pi }{6}+i\sin \frac{11\pi }{6}\right)\end{array}$

Therefore, the standard form of the roots,

  $\begin{array}{l}5\left(\cos \frac{\pi }{2}+i\sin \frac{\pi }{2}\right)=2.9240i\\ 5\left(\cos \frac{7\pi }{6}+i\sin \frac{7\pi }{6}\right)=-2.5323-1.4620i\\ 5\left(\cos \frac{11\pi }{6}+i\sin \frac{11\pi }{6}\right)=2.5323-1.4620i\end{array}$

c)

To determine

To graph: The cube roots of $-25i$

Explanation of Solution

Given information: The roots of $-25i$

Graph: The required graph of roots are given below,