Chapter 9.3, Problem 60AYU

To determine

To find: The fifth root of a graph complex number.

Answer to Problem 60AYU

The fifth root $-i$ is $cis\left(\frac{2\pi k}{5}\right)$

Explanation of Solution

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

Calculation:

  $\begin{array}{c}z=a+ib\\ =r\left(\cos a+i\sin b\right)\\ =r{e}^{i\theta }\sin \text{ce, }{e}^{i\theta }=\cos \theta +i\sin \theta \end{array}$ Where, $r=\sqrt{a^2+b^2}\text{ and }\theta {\text{=tan}}^{-1}\left(\frac{b}{a}\right)$

Now, $r{e}^{i\theta }=r{e}^{i\left(\theta +2\pi k\right)}\text{ , k}\in \mathbb{Z}$

For, The $n^{th}$ root:

  $\begin{array}{c}{\left(a+ib\right)}^{\frac{1}{n}}={\left(r{e}^{i\left(\theta +2\pi k\right)}\right)}^{\frac{1}{n}}\\ =r^{\frac{1}{n}}{e}^{i\left(\frac{\theta +2\pi k}{n}\right)}\end{array}$

For, Fifth root $k=0,1,2,3,4$

For, $k=0$

  $\begin{array}{c}{\left(a+ib\right)}^{\frac{1}{n}}=r^{\frac{1}{n}}{e}^{i\theta }\\ =r^{\frac{1}{n}}\left(\cos \theta +\sin \theta \right)\end{array}$

  $k=1$

  $\begin{array}{c}\left(a+ib\right)=r^{\frac{1}{n}}\left[e^{i\left(\frac{\theta +2\pi }{n}\right)}\right]\\ =r^{\frac{1}{n}}\left[\cos \left(\frac{\theta +2\pi }{n}\right)+\sin \left(\frac{\theta +2\pi }{n}\right)\right]\end{array}$

Then, ${\left(a+ib\right)}^{\frac{1}{n}}=r^{\frac{1}{n}}cis\left(\frac{\theta +2\pi k}{n}\right),\text{ k=0,1,}\mathrm{............}$

Now, for $-i$

  $a=0,b=-1$

Then,

  $\begin{array}{c}r=\sqrt{0^2+{\left(-1\right)}^2}\\ \theta ={\tan }^{-1}\left(\left|\frac{b}{a}\right|\right)\\ ={\tan }^{-1}\left(0\right)\\ =0\end{array}$

The fifth root $-i$ is, since, fifth root $n=5$

  $-i=cis\left(\frac{2\pi k}{5}\right),\text{ k = 0,1,2,3,4}$

For,

  $\begin{array}{l}k=0,\text{ 1}\\ k=1,cis\left(\frac{2\pi }{5}\right)\\ k=2,cis\left(\frac{4\pi }{5}\right)\\ k=3,cis\left(\frac{6\pi }{5}\right)\\ k=4,cis\left(\frac{8\pi }{5}\right)\\ \\ Where,cis\left(x\right)=\cos \left(x\right)+i\sin \left(x\right)\end{array}$