Answer
The power of the complex numbers in the rectangular form is $-64$.
Work Step by Step
Consider the given complex number to write in the polar form,
$z={{\left( \sqrt{3}-i \right)}^{6}}$ ...... (1)
For a complex number $z=x+iy$, the polar form is given by,
$z=r\left( \cos \theta +i\sin \theta \right)$ ...... (2)
Here, $x=r\cos \theta \ \text{ and }\ y=r\sin \theta $, dividing the value of y by x, to get,
$\tan \theta =\frac{y}{x}$ ...... (3)
Also, the value of r is called the moduli of the complex number, given by,
$\begin{align}
& r=\sqrt{{{x}^{2}}+{{y}^{2}}} \\
& =\sqrt{{{\left( \sqrt{3} \right)}^{2}}+{{\left( -1 \right)}^{2}}} \\
& =\sqrt{3+1} \\
& =2
\end{align}$
For any complex number $z=x+iy$, the sign of the value of x and y determine in which quadrant the value of $z=x+iy$ would lie.
If the value of x lies on the positive side and the value of y is positive, then the angle $\theta $ lies in the first quadrant having the value of $\theta $ as $\theta $.
If the value of x lies on the negative side and the value of y is positive, then the angle $\theta $ lies in the second quadrant having the value of $\theta $ as $\pi -\theta $.
If the value of x lies on the negative side and the value of y is negative, then the angle $\theta $ lies in the third quadrant having the value of $\theta $ as $\pi +\theta $.
If the value of x lies on the positive side and the value of y is negative, then the angle $\theta $ lies in the fourth quadrant having the value of $\theta $ as $2\pi -\theta $.
For the given complex number, using (1), (3) to get,
$\tan \theta =\frac{y}{x}=\frac{-1}{\sqrt{3}}=-\frac{1}{\sqrt{3}}$
For the given complex number,
$\theta =\frac{11\pi }{6}$
Use (2), to get,
$\left( \sqrt{3}-i \right)=2\left( \cos \frac{11\pi }{6}+i\sin \frac{11\pi }{6} \right)$
The polar form of the complex number is,
$z={{\left[ 2\left( \cos \frac{11\pi }{6}+i\sin \frac{11\pi }{6} \right) \right]}^{6}}$
Consider the given complex number in the polar form,
$z={{\left[ 2\left( \cos \frac{11\pi }{6}+i\sin \frac{11\pi }{6} \right) \right]}^{6}}$ ...... (4)
If n is a positive integer, then ${{z}^{n}}$ is,
$\begin{align}
& {{z}^{n}}={{\left[ r\left( \cos \theta +i\sin \theta \right) \right]}^{n}} \\
& ={{r}^{n}}\left( \cos n\theta +i\sin n\theta \right)
\end{align}$
That is.,
${{z}^{n}}={{r}^{n}}\left( \cos n\theta +i\sin n\theta \right)$
Now, for the given complex number, use (4) to get,
$\begin{align}
& z={{\left[ 2\left( \cos \frac{11\pi }{6}+i\sin \frac{11\pi }{6} \right) \right]}^{6}} \\
& z={{2}^{6}}\left( \cos 6\times \frac{11\pi }{6}+i\sin 6\times \frac{11\pi }{6} \right) \\
& z=64\left( \cos 11\pi +i\sin 11\pi \right) \\
\end{align}$
Simplifying it further, to get,
$\begin{align}
& z=64\left( -1+i0 \right) \\
& z=-64 \\
\end{align}$
The power of the complex numbers in the rectangular form is $-64$.
