Answer
$128+128i$
Work Step by Step
For converting the complex number in rectangular form $z=a+ib$, the polar form of the complex number is given by
$z=r\left( \cos \theta +i\sin \theta \right)$
Where
$\begin{align}
& r=\sqrt{{{a}^{2}}+{{b}^{2}}} \\
& \theta ={{\tan }^{-1}}\left( \frac{b}{a} \right)
\end{align}$
DeMoivre’s Theorem:
For any complex number $z=r\left( \cos \theta +i\sin \theta \right)$, if n is a positive integer $\left( z>0 \right)$ then,
$\begin{align}
& {{z}^{n}}={{\left[ r\left( \cos \theta +i\sin \theta \right) \right]}^{n}} \\
& ={{r}^{n}}\left( \cos n\theta +i\sin n\theta \right)
\end{align}$
So, at first we will convert $\left( -2-2i \right)$ in the polar form.
Here,
$\begin{align}
& r=\sqrt{{{\left( -2 \right)}^{2}}+{{\left( -2 \right)}^{2}}} \\
& =2\sqrt{2}
\end{align}$
Since, $\left( -2-2i \right)$ lies in ${{3}^{\text{rd}}}$ quadrant, we have
$\begin{align}
& \theta ={{\tan }^{-1}}\left( \frac{-2}{-2} \right) \\
& \theta ={{\tan }^{-1}}\left( 1 \right) \\
& \theta =\pi +\frac{\pi }{4} \\
& =\frac{5\pi }{4}
\end{align}$
So, $\left( -2-2i \right)=2\sqrt{2}\left( \cos \frac{5\pi }{4}+i\sin \frac{5\pi }{4} \right)$.
Therefore, ${{\left( \left( -2-2i \right) \right)}^{5}}={{\left[ 2\sqrt{2}\left( \cos \frac{5\pi }{4}+i\sin \frac{5\pi }{4} \right) \right]}^{5}}$.
So,
$\begin{align}
& {{\left[ 2\sqrt{2}\left( \cos \frac{5\pi }{4}+i\sin \frac{5\pi }{4} \right) \right]}^{5}}={{\left( 2\sqrt{2} \right)}^{5}}\left[ \cos 5\left( \frac{5\pi }{4} \right)+i\sin 5\left( \frac{5\pi }{4} \right) \right] \\
& =128\sqrt{2}\left( \cos \frac{25\pi }{4}+i\sin \frac{25\pi }{4} \right) \\
& =128\sqrt{2}\left\{ \cos \left( 6\pi +\frac{\pi }{4} \right)+i\sin \left( 6\pi +\frac{\pi }{4} \right) \right\} \\
& =128\sqrt{2}\left( \frac{1}{\sqrt{2}}+i\frac{1}{\sqrt{2}} \right)
\end{align}$
That is, $128+128i$.
Hence, ${{\left( -2-2i \right)}^{5}}=128+128i$.
