Chapter 8, Problem 6T

To determine

To evaluate: The cube roots of complex number $27i$ and sketch the cube roots obtained

in the complex plane.

Answer to Problem 6T

The cube roots ${\omega }_{°}$, ${\omega }_1$ and ${\omega }_2$ of complex number $27i$ are $\left(\frac{3\sqrt{3}}{2}+i\frac{3}{2}\right)$, $\left(\frac{3\sqrt{3}}{2}+i\frac{3}{2}\right)$ and $-3i$.

Explanation of Solution

Given: The complex number is $27i$.

Formula Used:

The polar form of a complex number is $z=r\left(\cos \theta +i\sin \theta \right)$

Then the $n$ distinct complex roots are ${\omega }_k=r^{\frac{1}{n}}\left(\cos \left(\frac{\theta +2\pi k}{n}\right)+i\sin \left(\frac{\theta +2\pi k}{n}\right)\right)$ for $k=0,1,2,3\cdots$

Calculation:

The polar form of a complex number is,

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

Where,

The number $r$ is the modulus of $z$.

$r=\left|z\right|=\sqrt{a^2+b^2}$

The $\theta$ is the argument of $z$.

The value of $r$ is $r=\sqrt{a^2+b^2}$

Substitute 0 for $a$ and 27 for $b$.

$\begin{array}{c}r=\sqrt{{\left(0\right)}^2+{\left(27\right)}^2}\\ =\sqrt{{\left(27\right)}^2}\\ =27\end{array}$

The value of $r$ is 27.

The value of $\theta$ is $\tan \theta =\frac{b}{a}$.

Substitute 0 for $a$ and 27 for $b$.

$\begin{array}{c}\tan \theta =\frac{27}{0}\\ =\infty \end{array}$

Possible value of $\theta$ is $\frac{\pi }{2}$

The complex number lies in first quadrant, so the value of $\theta$ is $\frac{\pi }{2}$.

The polar form of the complex number $z$ is

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

Substitute 27 for $r$ and $\frac{\pi }{2}$ for $\theta$.

$z=27\left(\cos \left(\frac{\pi }{2}\right)+i\sin \left(\frac{\pi }{2}\right)\right)$

The rule of $nth$ Roots of complex number is ${\omega }_k=r^{\frac{1}{n}}\left(\cos \left(\frac{\theta +2\pi k}{n}\right)+i\sin \left(\frac{\theta +2\pi k}{n}\right)\right)$.

Substitute 27 for $r$, $\left(\frac{\pi }{2}\right)$ for $\theta$, 3 for $n$ and 0 for $k$.

$\begin{array}{c}{\omega }_{°}={27}^{\frac{1}{3}}\left(\cos \frac{\frac{\pi }{2}+2\pi \left(0\right)}{3}+i\sin \frac{\frac{\pi }{2}+2\pi \left(0\right)}{3}\right)\\ =3\left(\cos \left(\frac{\pi }{6}\right)+i\sin \left(\frac{\pi }{6}\right)\right)\\ =3\left(\frac{\sqrt{3}}{2}+i\frac{1}{2}\right)\\ =\left(\frac{3\sqrt{3}}{2}+i\frac{3}{2}\right)\end{array}$

Substitute 27 for $r$, $\left(\frac{\pi }{2}\right)$ for $\theta$, 3 for $n$ and 1 for $k$.

$\begin{array}{c}{\omega }_1={27}^{\frac{1}{3}}\left(\cos \frac{\frac{\pi }{2}+2\pi \left(1\right)}{3}+i\sin \frac{\frac{\pi }{2}+2\pi \left(1\right)}{3}\right)\\ =3\left(\cos \left(\frac{5\pi }{6}\right)+i\sin \left(\frac{5\pi }{6}\right)\right)\\ =3\left(\frac{-\sqrt{3}}{2}+i\frac{1}{2}\right)\\ =-\frac{3\sqrt{3}}{2}+i\frac{3}{2}\end{array}$

Substitute 27 for $r$, $\left(\frac{\pi }{2}\right)$ for $\theta$, 3 for $n$ and 2 for $k$.

$\begin{array}{c}{\omega }_1={27}^{\frac{1}{3}}\left(\cos \frac{\frac{\pi }{2}+2\pi \left(2\right)}{3}+i\sin \frac{\frac{\pi }{2}+2\pi \left(2\right)}{3}\right)\\ =3\left(\cos \left(\frac{9\pi }{6}\right)+i\sin \left(\frac{9\pi }{6}\right)\right)\\ =3\left(0-i\right)\\ =-3i\end{array}$

Thus, the cube roots ${\omega }_{°}$, ${\omega }_1$ and ${\omega }_2$ of complex number $27i$ $\left(\frac{3\sqrt{3}}{2}+i\frac{3}{2}\right)$, $\left(\frac{-3\sqrt{3}}{2}+i\frac{3}{2}\right)$ and $-3i$.

The below graph is of the given cube roots of the complex number $27i$ in complex plane.

Figure (1)