Chapter 6.6, Problem 115E

To determine

To find: The fourth roots of $256i$ .

Answer to Problem 115E

The required roots are as follows:

  $4\left(\cos \frac{\pi }{8}+i\sin \frac{\pi }{8}\right),4\left(\cos \frac{5\pi }{8}+i\sin \frac{5\pi }{8}\right),4\left(\cos \frac{9\pi }{8}+i\sin \frac{9\pi }{8}\right)\&4\left(\cos \frac{13\pi }{8}+i\sin \frac{13\pi }{8}\right)$

Explanation of Solution

Given information: The complex number is $256i$ .

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: A complex number $z=a+bi$ can be rewritten as $z=r\left(\cos \theta +i\sin \theta \right)$

Hence,

  $\begin{array}{l}r=\sqrt{a^2+b^2}\\ =\sqrt{0^2+{\left(256\right)}^2}\\ =256\end{array}$

Now,

  $\begin{array}{l}\tan \theta =\frac{b}{a}\\ =\frac{256}{0}\\ =\infty \end{array}$

  $\therefore \theta =\frac{\pi }{2}$ lies in I quadrant.

Now,

  $256i=256\left(\cos \frac{\pi }{2}+i\sin \frac{\pi }{2}\right)$

Apply the $n^{th}$ roots formula to obtain the 4 roots of the complex number.

  $\begin{array}{l}\sqrt[4]{256}\left(\cos \frac{\frac{\pi }{2}+2\pi k}{4}+i\sin \frac{\frac{\pi }{2}+2\pi k}{4}\right)=4\left(\cos \frac{\pi +4\pi k}{8}+i\sin \frac{\pi +4\pi k}{8}\right)\\ 4\left(\cos \frac{\pi +4\pi \left(0\right)}{8}+i\sin \frac{\pi +4\pi \left(0\right)}{8}\right)=4\left(\cos \frac{\pi }{8}+i\sin \frac{\pi }{8}\right)\\ 4\left(\cos \frac{\pi +4\pi \left(1\right)}{8}+i\sin \frac{\pi +4\pi \left(1\right)}{8}\right)=4\left(\cos \frac{5\pi }{8}+i\sin \frac{5\pi }{8}\right)\\ 4\left(\cos \frac{\pi +4\pi \left(2\right)}{8}+i\sin \frac{\pi +4\pi \left(2\right)}{8}\right)=4\left(\cos \frac{9\pi }{8}+i\sin \frac{9\pi }{8}\right)\\ 4\left(\cos \frac{\pi +4\pi \left(3\right)}{8}+i\sin \frac{\pi +4\pi \left(3\right)}{8}\right)=4\left(\cos \frac{13\pi }{8}+i\sin \frac{13\pi }{8}\right)\end{array}$

To determine

To find: The fourth roots of $256i$ .

Answer to Problem 115E

The standard forms of the roots are as follows:

  $\left(3.6956+1.5308i\right),\left(-1.5308+3.6956i\right),\left(-3.6956-1.5308i\right)\&\left(1.5308-3.6956i\right)$

Explanation of Solution

Given information: The complex number is $256i$ .

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: A complex number $z=a+bi$ can be rewritten as $z=r\left(\cos \theta +i\sin \theta \right)$

Hence,

  $\begin{array}{l}r=\sqrt{a^2+b^2}\\ =\sqrt{0^2+{\left(256\right)}^2}\\ =256\end{array}$

Now,

  $\begin{array}{l}\tan \theta =\frac{b}{a}\\ =\frac{256}{0}\\ =\infty \end{array}$

  $\therefore \theta =\frac{\pi }{2}$ lies in I quadrant.

Now,

  $256i=256\left(\cos \frac{\pi }{2}+i\sin \frac{\pi }{2}\right)$

Apply the $n^{th}$ roots formula to obtain the 4 roots of the complex number.

  $\begin{array}{l}\sqrt[4]{256}\left(\cos \frac{\frac{\pi }{2}+2\pi k}{4}+i\sin \frac{\frac{\pi }{2}+2\pi k}{4}\right)=4\left(\cos \frac{\pi +4\pi k}{8}+i\sin \frac{\pi +4\pi k}{8}\right)\\ 4\left(\cos \frac{\pi +4\pi \left(0\right)}{8}+i\sin \frac{\pi +4\pi \left(0\right)}{8}\right)=4\left(\cos \frac{\pi }{8}+i\sin \frac{\pi }{8}\right)\\ 4\left(\cos \frac{\pi +4\pi \left(1\right)}{8}+i\sin \frac{\pi +4\pi \left(1\right)}{8}\right)=4\left(\cos \frac{5\pi }{8}+i\sin \frac{5\pi }{8}\right)\\ 4\left(\cos \frac{\pi +4\pi \left(2\right)}{8}+i\sin \frac{\pi +4\pi \left(2\right)}{8}\right)=4\left(\cos \frac{9\pi }{8}+i\sin \frac{9\pi }{8}\right)\\ 4\left(\cos \frac{\pi +4\pi \left(3\right)}{8}+i\sin \frac{\pi +4\pi \left(3\right)}{8}\right)=4\left(\cos \frac{13\pi }{8}+i\sin \frac{13\pi }{8}\right)\end{array}$

The standard forms of the roots are as follows:

  $\begin{array}{l}4\left(\cos \frac{\pi }{8}+i\sin \frac{\pi }{8}\right)=\left(3.6956+1.5308i\right)\\ 4\left(\cos \frac{5\pi }{8}+i\sin \frac{5\pi }{8}\right)=\left(-1.5308+3.6956i\right)\\ 4\left(\cos \frac{9\pi }{8}+i\sin \frac{9\pi }{8}\right)=\left(-3.6956-1.5308i\right)\\ 4\left(\cos \frac{13\pi }{8}+i\sin \frac{13\pi }{8}\right)=\left(1.5308-3.6956i\right)\end{array}$

To determine

To graph: The roots of $256i$ .

Explanation of Solution

Given information: The complex number $256i$ .

Graph: The graphical representation of the roots are given below,