Answer
The complex fourth roots of the given complex numbers in rectangular form are,
${{z}_{0}}=\frac{3}{2}+i\frac{3\sqrt{3}}{2},{{z}_{1}}=-\frac{3\sqrt{3}}{2}+\frac{3}{2}i,\text{ }{{z}_{2}}=-\frac{3}{2}-\frac{3\sqrt{3}}{2}i\text{ and }{{z}_{3}}=\frac{3\sqrt{3}}{2}-\frac{3}{2}i$.
Work Step by Step
Consider the given complex number,
$81\left( \cos \frac{4\pi }{3}+i\cos \frac{4\pi }{3} \right)$
For any complex number of the type,
$z=r\cos \theta +i\sin \theta $
If n is a positive integer, then ${{z}^{n}}$ is,
${{z}_{k}}=\sqrt[n]{r}\left[ \cos \left( \frac{\theta +2\pi k}{n} \right)+i\sin \left( \frac{\theta +2\pi k}{n} \right) \right]$ ...... (1)
Now, for the given complex number, use (1) to get,
$\begin{align}
& {{z}_{k}}=\sqrt[n]{r}\left[ \cos \left( \frac{\theta +2\pi k}{n} \right)+i\sin \left( \frac{\theta +2\pi k}{n} \right) \right] \\
& {{z}_{0}}=\sqrt[4]{81}\left[ \cos \frac{\frac{4\pi }{3}+2\pi \times 0}{4}+i\sin \frac{\frac{4\pi }{3}+2\pi \times 0}{4} \right] \\
& {{z}_{0}}=3\left[ \cos \frac{\pi }{3}+i\sin \frac{\pi }{3} \right] \\
& {{z}_{0}}=\frac{3}{2}+i\frac{3\sqrt{3}}{2} \\
\end{align}$
Now, for the other roots, substitute the value of $k=0,1,2,3...$
That is,
$\begin{align}
& {{z}_{k}}=\sqrt[n]{r}\left[ \cos \left( \frac{\theta +2\pi k}{n} \right)+i\sin \left( \frac{\theta +2\pi k}{n} \right) \right] \\
& {{z}_{1}}=\sqrt[4]{81}\left[ \cos \frac{\frac{4\pi }{3}+2\pi \times 1}{4}+i\sin \frac{\frac{4\pi }{3}+2\pi \times 1}{4} \right] \\
& {{z}_{1}}=3\left[ \cos \frac{5\pi }{6}+i\sin \frac{5\pi }{6} \right] \\
& {{z}_{1}}=-\frac{3\sqrt{3}}{2}+\frac{3}{2}i \\
\end{align}$
Similarly,
$\begin{align}
& {{z}_{k}}=\sqrt[n]{r}\left[ \cos \left( \frac{\theta +2\pi k}{n} \right)+i\sin \left( \frac{\theta +2\pi k}{n} \right) \right] \\
& {{z}_{2}}=\sqrt[4]{81}\left[ \cos \frac{\frac{4\pi }{3}+2\pi \times 2}{4}+i\sin \frac{\frac{4\pi }{3}+2\pi \times 2}{4} \right] \\
& {{z}_{2}}=3\left[ \cos \frac{4\pi }{3}+i\sin \frac{4\pi }{3} \right] \\
& {{z}_{2}}=-\frac{3}{2}-\frac{3\sqrt{3}}{2}i \\
\end{align}$
Also,
$\begin{align}
& {{z}_{k}}=\sqrt[n]{r}\left[ \cos \left( \frac{\theta +2\pi k}{n} \right)+i\sin \left( \frac{\theta +2\pi k}{n} \right) \right] \\
& {{z}_{3}}=\sqrt[4]{81}\left[ \cos \frac{\frac{4\pi }{3}+2\pi \times 3}{4}+i\sin \frac{\frac{4\pi }{3}+2\pi \times 3}{4} \right] \\
& {{z}_{3}}=3\left[ \cos \frac{11\pi }{6}+i\sin \frac{11\pi }{6} \right] \\
& {{z}_{3}}=\frac{3\sqrt{3}}{2}-\frac{3}{2}i \\
\end{align}$
The fourth roots of the complex numbers in the polar form are,
${{z}_{0}}=\frac{3}{2}+i\frac{3\sqrt{3}}{2},{{z}_{1}}=-\frac{3\sqrt{3}}{2}+\frac{3}{2}i,\text{ }{{z}_{2}}=-\frac{3}{2}-\frac{3\sqrt{3}}{2}i\text{ and }{{z}_{3}}=\frac{3\sqrt{3}}{2}-\frac{3}{2}i$.
