Chapter 6, Problem 125CR

To determine

To find: The cube root of $343$ .

Answer to Problem 125CR

The roots of the given complex number are as follows:

  $\begin{array}{l}7\left(\cos 0+i\sin 0\right)\\ 7\left(\cos \frac{2\pi }{3}+i\sin \frac{2\pi }{3}\right)\\ 7\left(\cos \frac{4\pi }{3}+i\sin \frac{4\pi }{3}\right)\end{array}$

Explanation of Solution

Given information: The number $343$

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 can be rewritten as $z=r\left(\cos \theta +i\sin \theta \right)$

Hence,

  $\begin{array}{l}r=\sqrt{a^2+b^2}\\ =\sqrt{{\left(343\right)}^2+0^2}\\ =343\end{array}$

Now,

  $\begin{array}{l}\tan \theta =\frac{b}{a}\\ =\frac{0}{343}\\ =0\end{array}$

  $\therefore \theta =0$ lies in I quadrant.

Now,

  $343=343\left(\cos 0+i\sin 0\right)$

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

  $\sqrt[3]{343}\left(\cos \frac{0+2\pi k}{3}+i\sin \frac{0+2\pi k}{3}\right)=7\left(\cos \frac{2\pi k}{3}+i\sin \frac{2\pi k}{3}\right)$

  $\begin{array}{l}\sqrt[3]{343}\left(\cos \frac{0+2\pi k}{3}+i\sin \frac{0+2\pi k}{3}\right)=7\left(\cos \frac{2\pi k}{3}+i\sin \frac{2\pi k}{3}\right)\\ 7\left(\cos \frac{2\pi \left(0\right)}{3}+i\sin \frac{2\pi \left(0\right)}{3}\right)=7\left(\cos 0+i\sin 0\right)\\ 7\left(\cos \frac{2\pi \left(1\right)}{3}+i\sin \frac{2\pi \left(1\right)}{3}\right)=7\left(\cos \frac{2\pi }{3}+i\sin \frac{2\pi }{3}\right)\\ 7\left(\cos \frac{2\pi \left(2\right)}{3}+i\sin \frac{2\pi \left(2\right)}{3}\right)=7\left(\cos \frac{4\pi }{3}+i\sin \frac{4\pi }{3}\right)\end{array}$

To determine

To find: standard form of the root.

Answer to Problem 125CR

The standard form of the roots are as follows:

  $\begin{array}{l}7\\ \left(-\frac{1}{2}+\frac{\sqrt{3}}{2}i\right)\\ 7\left(-\frac{1}{2}-\frac{\sqrt{3}}{2}i\right)\end{array}$

Explanation of Solution

Given information: The number $343$

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 can be rewritten as $z=r\left(\cos \theta +i\sin \theta \right)$

Hence,

  $\begin{array}{l}r=\sqrt{a^2+b^2}\\ =\sqrt{{\left(343\right)}^2+0^2}\\ =343\end{array}$

Now,

  $\begin{array}{l}\tan \theta =\frac{b}{a}\\ =\frac{0}{343}\\ =0\end{array}$

  $\therefore \theta =0$ lies in I quadrant.

Now,

  $343=343\left(\cos 0+i\sin 0\right)$

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

  $\sqrt[3]{343}\left(\cos \frac{0+2\pi k}{3}+i\sin \frac{0+2\pi k}{3}\right)=7\left(\cos \frac{2\pi k}{3}+i\sin \frac{2\pi k}{3}\right)$

  $\begin{array}{l}\sqrt[3]{343}\left(\cos \frac{0+2\pi k}{3}+i\sin \frac{0+2\pi k}{3}\right)=7\left(\cos \frac{2\pi k}{3}+i\sin \frac{2\pi k}{3}\right)\\ 7\left(\cos \frac{2\pi \left(0\right)}{3}+i\sin \frac{2\pi \left(0\right)}{3}\right)=7\left(\cos 0+i\sin 0\right)\\ 7\left(\cos \frac{2\pi \left(1\right)}{3}+i\sin \frac{2\pi \left(1\right)}{3}\right)=7\left(\cos \frac{2\pi }{3}+i\sin \frac{2\pi }{3}\right)\\ 7\left(\cos \frac{2\pi \left(2\right)}{3}+i\sin \frac{2\pi \left(2\right)}{3}\right)=7\left(\cos \frac{4\pi }{3}+i\sin \frac{4\pi }{3}\right)\end{array}$

By simplifying the roots further,

  $\begin{array}{l}7\left(\cos 0+i\sin 0\right)=7\\ 7\left(\cos \frac{2\pi }{3}+i\sin \frac{2\pi }{3}\right)=7\left(-\frac{1}{2}+\frac{\sqrt{3}}{2}i\right)\\ 7\left(\cos \frac{4\pi }{3}+i\sin \frac{4\pi }{3}\right)=7\left(-\frac{1}{2}-\frac{\sqrt{3}}{2}i\right)\end{array}$

To determine

To graph: The roots of $343$

Explanation of Solution

Given information: The number is $343$

Graph: The required graph represented below,