Chapter 6.6, Problem 114E

a)

To determine

To find: The cube roots of $-4\sqrt{2}\left(1-i\right)$

Answer to Problem 114E

The cube roots of $-4\sqrt{2}\left(1-i\right)$ :

  $\begin{array}{l}\Rightarrow 2\left(\cos \frac{\pi }{4}+i\sin \frac{\pi }{4}\right)\\ \Rightarrow 2\left(\cos \frac{11\pi }{12}+i\sin \frac{11\pi }{12}\right)\\ \Rightarrow 2\left(\cos \frac{19\pi }{12}+i\sin \frac{19\pi }{12}\right)\end{array}$

Explanation of Solution

Given information: The complex number $-4\sqrt{2}\left(1-i\right)$

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.

  $-4\sqrt{2}\left(1-i\right)=8\left(\cos \frac{3\pi }{4}+i\sin \frac{3\pi }{4}\right)$

  $\begin{array}{l}\sqrt[3]{8}\left(\cos \frac{\frac{3\pi }{4}+2\pi k}{3}+i\sin \frac{\frac{3\pi }{4}+2\pi k}{3}\right)=2\left(\cos \frac{3\pi +8\pi k}{12}+i\sin \frac{3\pi +8\pi k}{12}\right)\\ \therefore 2\left(\cos \frac{3\pi +8\pi \left(0\right)}{12}+i\sin \frac{3\pi +8\pi \left(0\right)}{12}\right)=2\left(\cos \frac{\pi }{4}+i\sin \frac{\pi }{4}\right)\\ \therefore 2\left(\cos \frac{3\pi +8\pi \left(1\right)}{12}+i\sin \frac{3\pi +8\pi \left(1\right)}{12}\right)=2\left(\cos \frac{11\pi }{12}+i\sin \frac{11\pi }{12}\right)\\ \therefore 2\left(\cos \frac{3\pi +8\pi \left(2\right)}{12}+i\sin \frac{3\pi +8\pi \left(2\right)}{12}\right)=2\left(\cos \frac{19\pi }{12}+i\sin \frac{19\pi }{12}\right)\end{array}$

b)

To determine

To find: The cube roots of $-4\sqrt{2}\left(1-i\right)$

Answer to Problem 114E

The roots of $-4\sqrt{2}\left(1-i\right)$ are in standard form:

  $\begin{array}{l}\Rightarrow \sqrt{2}+i\sqrt{2}\\ \Rightarrow -\frac{\sqrt{6}+\sqrt{2}-i\sqrt{6}+i\sqrt{2}}{2}\\ \Rightarrow \sqrt{2-\sqrt{3}}-i\sqrt{2+\sqrt{3}}\end{array}$

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.

  $-4\sqrt{2}\left(1-i\right)=8\left(\cos \frac{3\pi }{4}+i\sin \frac{3\pi }{4}\right)$

  $\begin{array}{l}\sqrt[3]{8}\left(\cos \frac{\frac{3\pi }{4}+2\pi k}{3}+i\sin \frac{\frac{3\pi }{4}+2\pi k}{3}\right)=2\left(\cos \frac{3\pi +8\pi k}{12}+i\sin \frac{3\pi +8\pi k}{12}\right)\\ \therefore 2\left(\cos \frac{3\pi +8\pi \left(0\right)}{12}+i\sin \frac{3\pi +8\pi \left(0\right)}{12}\right)=2\left(\cos \frac{\pi }{4}+i\sin \frac{\pi }{4}\right)\\ \therefore 2\left(\cos \frac{3\pi +8\pi \left(1\right)}{12}+i\sin \frac{3\pi +8\pi \left(1\right)}{12}\right)=2\left(\cos \frac{11\pi }{12}+i\sin \frac{11\pi }{12}\right)\\ \therefore 2\left(\cos \frac{3\pi +8\pi \left(2\right)}{12}+i\sin \frac{3\pi +8\pi \left(2\right)}{12}\right)=2\left(\cos \frac{19\pi }{12}+i\sin \frac{19\pi }{12}\right)\end{array}$

Therefore, the standard form of the roots,

  $\begin{array}{l}\Rightarrow 2\left(\cos \frac{\pi }{4}+i\sin \frac{\pi }{4}\right)=\sqrt{2}+i\sqrt{2}\\ \Rightarrow 2\left(\cos \frac{11\pi }{12}+i\sin \frac{11\pi }{12}\right)=-\frac{\sqrt{6}+\sqrt{2}-i\sqrt{6}+i\sqrt{2}}{2}\\ \Rightarrow 2\left(\cos \frac{19\pi }{12}+i\sin \frac{19\pi }{12}\right)=\sqrt{2-\sqrt{3}}-i\sqrt{2+\sqrt{3}}\end{array}$

c)

To determine

To graph: The cube roots of $-4\sqrt{2}\left(1-i\right)$

Explanation of Solution

Given information: The roots of $-4\sqrt{2}\left(1-i\right)$

Graph: The required graph of roots are given below,