Chapter 6, Problem 112RE

To determine

To find: The square root of the given complex number.

Answer to Problem 112RE

The square root is $\pm \left(0.3660+i1.3660\right)$ .

Explanation of Solution

Given:

The complex number is $\sqrt{3}-i$ .

Calculation:

For any complex number $z=r\left(\cos \theta +i\sin \theta \right)$ , the nth root is given by,

  $\sqrt[n]{r}\left(\cos \frac{\theta +2\pi k}{n}+i\sin \frac{\theta +2\pi k}{n}\right),k=0,1,2,\dots ,n-1$

So, to find the square root of $\sqrt{3}-i$ , let $z=\sqrt{3}-i$ and convert it into the form $r\left(\cos \theta +i\sin \theta \right)$ .

  $\begin{array}{c}r=\sqrt{a^2+b^2}\\ =\sqrt{{\left(\sqrt{3}\right)}^2+{\left(-1\right)}^2}\\ =\sqrt{3+1}\\ =2\end{array}$

  $\begin{array}{c}\tan \theta =\frac{b}{a}\\ =-\frac{1}{\sqrt{3}}\\ =\frac{11\pi }{6}\end{array}$

So, $\sqrt{3}-i=2\left(\cos \frac{11\pi }{6}+i\sin \frac{11\pi }{6}\right)$ and the square root is calculated as shown below:

  $\sqrt[2]{2}\left(\cos \frac{\frac{11\pi }{6}+2\pi k}{2}+i\sin \frac{\frac{11\pi }{6}+2\pi k}{2}\right)=\sqrt{2}\left(\cos \frac{11\pi +12\pi k}{12}+i\sin \frac{11\pi +12\pi k}{12}\right)$

For $k=0$ ,

  $\begin{array}{c}\sqrt{2}\left(\cos \frac{11\pi +12\pi \left(0\right)}{12}+i\sin \frac{11\pi +12\pi \left(0\right)}{12}\right)=\sqrt{2}\left(\cos \frac{11\pi }{12}+i\sin \frac{11\pi }{12}\right)\\ =\sqrt{2}\left(0.2588+i0.9659\right)\\ =0.3660+i1.3660\end{array}$

For $k=1$ ,

  $\begin{array}{c}\sqrt{2}\left(\cos \frac{11\pi +12\pi \left(1\right)}{12}+i\sin \frac{11\pi +12\pi \left(1\right)}{12}\right)=\sqrt{2}\left(\cos \frac{23\pi }{12}+i\sin \frac{23\pi }{12}\right)\\ =\sqrt{2}\left(-0.2588-i0.9659\right)\\ =-0.3660-i1.3660\end{array}$

So, the square root of the given complex number is $\pm \left(0.3660+i1.3660\right)$ .