Chapter 6.6, Problem 121E

a)

To determine

To find: The fifth roots of $128\left(-1+i\right)$

Answer to Problem 121E

The fifth roots of $128\left(-1+i\right)$ are as follows:

  $\begin{array}{l}\Rightarrow 2\sqrt{2}\left(\cos \frac{3\pi }{20}+i\sin \frac{3\pi }{20}\right)\\ \Rightarrow 2\sqrt{2}\left(\cos \frac{11\pi }{20}+i\sin \frac{11\pi }{20}\right)\\ \Rightarrow 2\sqrt{2}\left(\cos \frac{19\pi }{20}+i\sin \frac{19\pi }{20}\right)\\ \Rightarrow 2\sqrt{2}\left(\cos \frac{27\pi }{20}+i\sin \frac{27\pi }{20}\right)\\ \Rightarrow 2\sqrt{2}\left(\cos \frac{7\pi }{4}+i\sin \frac{7\pi }{4}\right)\end{array}$

Explanation of Solution

Given information: The complex number $128\left(-1+i\right)$

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.

  $128\left(-1+i\right)=128\sqrt{2}\left(\cos \frac{3\pi }{4}+i\sin \frac{3\pi }{4}\right)$

  $\begin{array}{l}{\left(128\sqrt{2}\right)}^{\frac{1}{5}}\left(\cos \frac{\frac{3\pi }{4}+2\pi k}{5}+i\sin \frac{\frac{3\pi }{4}+2\pi k}{5}\right)=2\sqrt{2}\left(\cos \frac{3\pi +8\pi k}{20}+i\sin \frac{3\pi +8\pi k}{20}\right)\\ \therefore 2\sqrt{2}\left(\cos \frac{3\pi +8\pi \left(0\right)}{20}+i\sin \frac{3\pi +8\pi \left(0\right)}{20}\right)=2\sqrt{2}\left(\cos \frac{3\pi }{20}+i\sin \frac{3\pi }{20}\right)\\ \therefore 2\sqrt{2}\left(\cos \frac{3\pi +8\pi \left(1\right)}{20}+i\sin \frac{3\pi +8\pi \left(1\right)}{20}\right)=2\sqrt{2}\left(\cos \frac{11\pi }{20}+i\sin \frac{11\pi }{20}\right)\\ \therefore 2\sqrt{2}\left(\cos \frac{3\pi +8\pi \left(2\right)}{20}+i\sin \frac{3\pi +8\pi \left(2\right)}{20}\right)=2\sqrt{2}\left(\cos \frac{19\pi }{20}+i\sin \frac{19\pi }{20}\right)\\ \therefore 2\sqrt{2}\left(\cos \frac{3\pi +8\pi \left(3\right)}{20}+i\sin \frac{3\pi +8\pi \left(3\right)}{20}\right)=2\sqrt{2}\left(\cos \frac{27\pi }{20}+i\sin \frac{27\pi }{20}\right)\\ \therefore 2\sqrt{2}\left(\cos \frac{3\pi +8\pi \left(4\right)}{20}+i\sin \frac{3\pi +8\pi \left(4\right)}{20}\right)=2\sqrt{2}\left(\cos \frac{7\pi }{4}+i\sin \frac{7\pi }{4}\right)\end{array}$

b)

To determine

To find: The complex number $128\left(-1+i\right)$

Answer to Problem 121E

The standard forms of fourth roots of $128\left(-1+i\right)$ are as follows:

  $\begin{array}{l}\Rightarrow 2.52+1.28i\\ \Rightarrow -0.44+2.79i\\ \Rightarrow -2.79+0.44i\\ \Rightarrow -1.28-2.52i\\ \Rightarrow 2-2i\end{array}$

Explanation of Solution

Given information: The complex number $128\left(-1+i\right)$

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.

  $128\left(-1+i\right)=128\sqrt{2}\left(\cos \frac{3\pi }{4}+i\sin \frac{3\pi }{4}\right)$

  $\begin{array}{l}{\left(128\sqrt{2}\right)}^{\frac{1}{5}}\left(\cos \frac{\frac{3\pi }{4}+2\pi k}{5}+i\sin \frac{\frac{3\pi }{4}+2\pi k}{5}\right)=2\sqrt{2}\left(\cos \frac{3\pi +8\pi k}{20}+i\sin \frac{3\pi +8\pi k}{20}\right)\\ \therefore 2\sqrt{2}\left(\cos \frac{3\pi +8\pi \left(0\right)}{20}+i\sin \frac{3\pi +8\pi \left(0\right)}{20}\right)=2\sqrt{2}\left(\cos \frac{3\pi }{20}+i\sin \frac{3\pi }{20}\right)\\ \therefore 2\sqrt{2}\left(\cos \frac{3\pi +8\pi \left(1\right)}{20}+i\sin \frac{3\pi +8\pi \left(1\right)}{20}\right)=2\sqrt{2}\left(\cos \frac{11\pi }{20}+i\sin \frac{11\pi }{20}\right)\\ \therefore 2\sqrt{2}\left(\cos \frac{3\pi +8\pi \left(2\right)}{20}+i\sin \frac{3\pi +8\pi \left(2\right)}{20}\right)=2\sqrt{2}\left(\cos \frac{19\pi }{20}+i\sin \frac{19\pi }{20}\right)\\ \therefore 2\sqrt{2}\left(\cos \frac{3\pi +8\pi \left(3\right)}{20}+i\sin \frac{3\pi +8\pi \left(3\right)}{20}\right)=2\sqrt{2}\left(\cos \frac{27\pi }{20}+i\sin \frac{27\pi }{20}\right)\\ \therefore 2\sqrt{2}\left(\cos \frac{3\pi +8\pi \left(4\right)}{20}+i\sin \frac{3\pi +8\pi \left(4\right)}{20}\right)=2\sqrt{2}\left(\cos \frac{7\pi }{4}+i\sin \frac{7\pi }{4}\right)\end{array}$

Standard forms of the roots are as follows:

  $\begin{array}{l}\Rightarrow 2\sqrt{2}\left(\cos \frac{3\pi }{20}+i\sin \frac{3\pi }{20}\right)=2.52+1.28i\\ \Rightarrow 2\sqrt{2}\left(\cos \frac{11\pi }{20}+i\sin \frac{11\pi }{20}\right)=-0.44+2.79i\\ \Rightarrow 2\sqrt{2}\left(\cos \frac{19\pi }{20}+i\sin \frac{19\pi }{20}\right)=-2.79+0.44i\\ \Rightarrow 2\sqrt{2}\left(\cos \frac{27\pi }{20}+i\sin \frac{27\pi }{20}\right)=-1.28-2.52i\\ \Rightarrow 2\sqrt{2}\left(\cos \frac{7\pi }{4}+i\sin \frac{7\pi }{4}\right)=2-2i\end{array}$

c)

To determine

To graph: The fourth roots of $128\left(-1+i\right)$

Explanation of Solution

Given information: The fourth roots of $128\left(-1+i\right)$

Graph: The required graph of roots are given below,