Answer
The power of the complex number in the rectangular form is $z=-4+4i$.
Work Step by Step
Consider the given complex number and write it in the polar form,
$z={{\left( 1-i \right)}^{5}}$
For a complex number ${{z}_{1}}=x+iy=1-i$, the polar form is given by,
${{z}_{1}}=r\left( \cos \theta +i\sin \theta \right)$
We have, $x=1=r\cos \theta \,y=-1=r\sin \theta $, where
$\begin{align}
& r=\sqrt{{{x}^{2}}+{{y}^{2}}} \\
& r=\sqrt{{{\left( 1 \right)}^{2}}+{{\left( -1 \right)}^{2}}} \\
& r=\sqrt{1+1} \\
& r=\sqrt{2} \\
\end{align}$
$r=\sqrt{2}$
Therefore
$\begin{align}
& 1=\sqrt{2}\cos \theta \,-1=\sqrt{2}\sin \theta \\
& i.e,\frac{1}{\sqrt{2}}=\cos \theta \,-\frac{1}{\sqrt{2}}=\sin \theta \\
\end{align}$
Which gives $\theta =-\frac{\pi }{4}$
${{z}_{1}}=\sqrt{2}\left( \cos (-\frac{\pi }{4})+i\sin (-\frac{\pi }{4}) \right)$
The polar form of the complex number is $z={{\left[ \sqrt{2}\left( \cos (-\frac{\pi }{4})+i\sin (-\frac{\pi }{4}) \right) \right]}^{5}}$
Therefore
$\begin{align}
& z={{\left[ \sqrt{2}\left( \cos (-\frac{\pi }{4})+i\sin (-\frac{\pi }{4}) \right) \right]}^{5}} \\
& z={{\left( \sqrt{2} \right)}^{5}}\left( \cos 5\times (-\frac{\pi }{4})+i\sin 5\times (-\frac{\pi }{4}) \right) \\
& z=4\sqrt{2}\left( \cos (-\frac{5\pi }{4})+i\sin (-\frac{5\pi }{4}) \right) \\
& z=4\sqrt{2}\left( \cos \frac{5\pi }{4}-i\sin \frac{5\pi }{4} \right) \\
\end{align}$
Simplify it further to get,
$\begin{align}
& z=4\sqrt{2}\left( -\frac{1}{\sqrt{2}}+i\frac{1}{\sqrt{2}} \right) \\
& z=-4+4i \\
\end{align}$
The complex number in the rectangular form is $-4+4i$
