Chapter 6.6, Problem 50E

Solution

To find: The fifth roots of the given complex number.

The fifth roots of the given complex number are $\sqrt[5]{2}\left(\cos \frac{\pi }{15}+i\sin \frac{\pi }{15}\right)$ , $\sqrt[5]{2}\left(\cos \frac{7\pi }{15}+i\sin \frac{7\pi }{15}\right)$ , $\sqrt[5]{2}\left(\cos \frac{13\pi }{15}+i\sin \frac{13\pi }{15}\right)$ , $\sqrt[5]{2}\left(\cos \frac{19\pi }{15}+i\sin \frac{19\pi }{15}\right)$ and $\frac{\left(1-\sqrt{3}i\right)}{\sqrt[5]{16}}$ .

Given information:

The complex number is $1+\sqrt{3}i$ .

Formula used:

The polar form of a complex number $z=a+bi$ is given by,

  $z=r\left(\cos \theta +i\sin \theta \right)$

Here, $r=\sqrt{a^2+b^2}$ and $\tan \theta =\frac{b}{a}$ .

The $n\text{th}$ roots of a complex number $z=r\left(\cos \theta +i\sin \theta \right)$ which are $n$ distinct complex numbers are given by,

  $\sqrt[n]{r}\left(\cos \frac{\theta +2\pi k}{n}+i\sin \frac{\theta +2\pi k}{n}\right)$

Here, $k=0,1,2,\mathrm{...},n-1$

Calculation:

From the given complex number,

  $\begin{array}{c}a=1\\ b=\sqrt{3}\end{array}$

Substitute $1$ for $a$ and $\sqrt{3}$ for $b$ in equation $r=\sqrt{a^2+b^2}$ to find modulus $r$ .

  $\begin{array}{c}r=\sqrt{{\left(1\right)}^2+{\left(\sqrt{3}\right)}^2}\\ =\sqrt{4}\\ =2\end{array}$

Substitute $1$ for $a$ and $\sqrt{3}$ for $b$ in equation $\tan \theta =\frac{b}{a}$ to find argument.

  $\begin{array}{c}\tan \theta =\frac{\sqrt{3}}{1}\\ \theta ={\tan }^{-1}\left(\sqrt{3}\right)\\ =\frac{\pi }{3}\end{array}$

Substitute $2$ for $r$ and $\frac{\pi }{3}$ for $\theta$ in equation $z=r\left(\cos \theta +i\sin \theta \right)$ .

  $z=2\left(\cos \frac{\pi }{3}+i\sin \frac{\pi }{3}\right)$

From the above complex number,

  $\begin{array}{c}r=2\\ \theta =\frac{\pi }{3}\\ n=5\end{array}$

Use $k=0,1,2,3\text{and}4$ to determine the cube roots.

Substitute $2$ for $r$ , $\frac{\pi }{3}$ for $\theta$ , $5$ for $n$ , and $0$ for $k$ in expression $\sqrt[n]{r}\left(\cos \frac{\theta +2\pi k}{n}+i\sin \frac{\theta +2\pi k}{n}\right)$ to find first root.

  $\begin{array}{c}{z}_1=\sqrt[5]{2}\left(\cos \frac{\frac{\pi }{3}+2\pi \cdot 0}{5}+i\sin \frac{\frac{\pi }{3}+2\pi \cdot 0}{5}\right)\\ =\sqrt[5]{2}\left(\cos \frac{\pi }{15}+i\sin \frac{\pi }{15}\right)\end{array}$

Substitute $2$ for $r$ , $\frac{\pi }{3}$ for $\theta$ , $5$ for $n$ , and $1$ for $k$ in expression $\sqrt[n]{r}\left(\cos \frac{\theta +2\pi k}{n}+i\sin \frac{\theta +2\pi k}{n}\right)$ to find second root.

  $\begin{array}{c}{z}_2=\sqrt[5]{2}\left(\cos \frac{\frac{\pi }{3}+2\pi \cdot 1}{5}+i\sin \frac{\frac{\pi }{3}+2\pi \cdot 1}{5}\right)\\ =\sqrt[5]{2}\left(\cos \frac{7\pi }{15}+i\sin \frac{7\pi }{15}\right)\end{array}$

Substitute $2$ for $r$ , $\frac{\pi }{3}$ for $\theta$ , $5$ for $n$ , and $2$ for $k$ in expression $\sqrt[n]{r}\left(\cos \frac{\theta +2\pi k}{n}+i\sin \frac{\theta +2\pi k}{n}\right)$ to find third root.

  $\begin{array}{c}{z}_3=\sqrt[5]{2}\left(\cos \frac{\frac{\pi }{3}+2\pi \cdot 2}{5}+i\sin \frac{\frac{\pi }{3}+2\pi \cdot 2}{5}\right)\\ =\sqrt[5]{2}\left(\cos \frac{13\pi }{15}+i\sin \frac{13\pi }{15}\right)\end{array}$

Substitute $2$ for $r$ , $\frac{\pi }{3}$ for $\theta$ , $5$ for $n$ , and $3$ for $k$ in expression $\sqrt[n]{r}\left(\cos \frac{\theta +2\pi k}{n}+i\sin \frac{\theta +2\pi k}{n}\right)$ to find fourth root.

  $\begin{array}{c}{z}_4=\sqrt[5]{2}\left(\cos \frac{\frac{\pi }{3}+2\pi \cdot 3}{5}+i\sin \frac{\frac{\pi }{3}+2\pi \cdot 3}{5}\right)\\ =\sqrt[5]{2}\left(\cos \frac{19\pi }{15}+i\sin \frac{19\pi }{15}\right)\end{array}$

Substitute $2$ for $r$ , $\frac{\pi }{3}$ for $\theta$ , $5$ for $n$ , and $4$ for $k$ in expression $\sqrt[n]{r}\left(\cos \frac{\theta +2\pi k}{n}+i\sin \frac{\theta +2\pi k}{n}\right)$ to find fifth root.

  $\begin{array}{c}{z}_5=\sqrt[5]{2}\left(\cos \frac{\frac{\pi }{3}+2\pi \cdot 4}{5}+i\sin \frac{\frac{\pi }{3}+2\pi \cdot 4}{5}\right)\\ =\sqrt[5]{2}\left(\cos \frac{5\pi }{3}+i\sin \frac{5\pi }{3}\right)\\ =\frac{\left(1-\sqrt{3}i\right)}{\sqrt[5]{16}}\end{array}$

Therefore, the fifth roots of the given complex number are $\sqrt[5]{2}\left(\cos \frac{\pi }{15}+i\sin \frac{\pi }{15}\right)$ , $\sqrt[5]{2}\left(\cos \frac{7\pi }{15}+i\sin \frac{7\pi }{15}\right)$ , $\sqrt[5]{2}\left(\cos \frac{13\pi }{15}+i\sin \frac{13\pi }{15}\right)$ , $\sqrt[5]{2}\left(\cos \frac{19\pi }{15}+i\sin \frac{19\pi }{15}\right)$ and $\frac{\left(1-\sqrt{3}i\right)}{\sqrt[5]{16}}$ .