Chapter 8.3, Problem 81E

To determine

To calculate: To find the indicated roots and graph the roots in the complex plane

Answer to Problem 81E

The square roots are ${\omega }_0=2\sqrt{2}\left[\cos \frac{\pi }{12}+i\sin \frac{\pi }{12}\right]$ and ${\omega }_1=2\sqrt{2}\left[\cos \frac{13\pi }{12}+i\sin \frac{13\pi }{12}\right]$

Explanation of Solution

Given information: The square roots of $4\sqrt{3}+4i$

Formula Used:

Complex number is a number that can be expressed in the form of $a+bi$ , where $a$ and $b$ are real numbers

  $z=a+bi$

If $z=r\left(\cos \theta +isin\theta \right)$ and $n$ is a positive integer, then $z$ has the $n$ distinct $n^{th}$ roots

  ${\omega }_k=r^{\frac{1}{n}}\left[\cos \left(\frac{\theta +2\pi k}{n}\right)+i\sin \left(\frac{\theta +2\pi k}{n}\right)\right]$ for $k=0,1,2,\mathrm{3....},n-1$

Calculation:

Complex number is given as

  $z=4\sqrt{3}+4i$

Polar form of complex number is

  $z=r\left(\cos \theta +isin\theta \right)--(1)$

Calculating the value of $r$ :

  $\begin{array}{l}r=\sqrt{{\left(4\sqrt{3}\right)}^2+{\left(4\right)}^2}\\ r=\sqrt{48+16}\\ r=\sqrt{64}\\ r=8\end{array}$

Calculating the value of ${\theta }_1$

  $\begin{array}{l}\tan \theta =\frac{4}{4\sqrt{3}}\\ \tan \theta =\frac{1}{\sqrt{3}}\\ \theta ={\tan }^{-1}\frac{1}{\sqrt{3}}\\ \theta =\frac{\pi }{6}\end{array}$

Hence, complex number in polar form is

  $z=8\left(\cos \frac{\pi }{6}+isin\frac{\pi }{6}\right)$

In order to find square roots of above complex number, apply nth Rule of Complex numbers

  ${\omega }_k=r^{\frac{1}{n}}\left[\cos \left(\frac{\theta +2\pi k}{n}\right)+i\sin \left(\frac{\theta +2\pi k}{n}\right)\right]$

Here, $n=2$ and $z=8\left(\cos \frac{\pi }{6}+isin\frac{\pi }{6}\right)$

Thus,

  $\begin{array}{l}{\omega }_k=8^{\frac{1}{2}}\left[\cos \left(\frac{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$6$}\right.+2\pi k}{2}\right)+i\sin \left(\frac{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$6$}\right.+2\pi k}{2}\right)\right]\\ {\omega }_k=2\sqrt{2}\left[\cos \left(\frac{\pi }{12}+\pi k\right)+i\sin \left(\frac{\pi }{12}+\pi k\right)\right]\end{array}$

Now, the value of $k$ is from $0$ to $n-1$

Thus,

  $\begin{array}{l}0\le k\le 2-1\\ 0\le k\le 1\end{array}$

Hence, $k=0,1$

When $k=0$

  $\begin{array}{l}{\omega }_0=2\sqrt{2}\left[\cos \left(\frac{\pi }{12}+\pi \left(0\right)\right)+i\sin \left(\frac{\pi }{12}+\pi \left(0\right)\right)\right]\\ {\omega }_0=2\sqrt{2}\left[\cos \frac{\pi }{12}+i\sin \frac{\pi }{12}\right]\end{array}$

When $k=1$

  $\begin{array}{l}{\omega }_1=2\sqrt{2}\left[\cos \left(\frac{\pi }{12}+\pi \left(1\right)\right)+i\sin \left(\frac{\pi }{12}+\pi \left(1\right)\right)\right]\\ {\omega }_1=2\sqrt{2}\left[\cos \frac{13\pi }{12}+i\sin \frac{13\pi }{12}\right]\end{array}$

Now, plotting the above points on the complex plane.

  

Conclusion:

Hence, square roots are ${\omega }_0=2\sqrt{2}\left[\cos \frac{\pi }{12}+i\sin \frac{\pi }{12}\right]$ and ${\omega }_1=2\sqrt{2}\left[\cos \frac{13\pi }{12}+i\sin \frac{13\pi }{12}\right]$