Answer
$-2-2\sqrt{3}i$
Work Step by Step
Conversion of a complex number from rectangular form to polar form:
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 $1-i\sqrt{3}$ in the polar form.
Here,
$\begin{align}
& r=\sqrt{{{1}^{2}}+{{\left( -\sqrt{3} \right)}^{2}}} \\
& =2
\end{align}$
And,
$\begin{align}
& \theta ={{\tan }^{-1}}\left( -\frac{\sqrt{3}}{1} \right) \\
& \theta ={{\tan }^{-1}}\left( -\sqrt{3} \right) \\
& \theta =300{}^\circ
\end{align}$
So, $1-i\sqrt{3}=2\left( \cos 300{}^\circ +i\sin 300{}^\circ \right)$.
Therefore, ${{\left( 1-i\sqrt{3} \right)}^{2}}={{\left[ 2\left( \cos 300{}^\circ +i\sin 300{}^\circ \right) \right]}^{2}}$.
So,
$\begin{align}
& {{\left[ 2\left( \cos 300{}^\circ +i\sin 300{}^\circ \right) \right]}^{2}}={{2}^{2}}\left[ \cos 2\left( 300{}^\circ \right)+i\sin 2\left( 300{}^\circ \right) \right] \\
& =4\left( \cos 600{}^\circ +i\sin 600{}^\circ \right) \\
& =4\left\{ \cos \left( 720-120 \right){}^\circ +i\sin \left( 720-120 \right){}^\circ \right\} \\
& =4\left( -\frac{1}{2}-i\frac{\sqrt{3}}{2} \right)
\end{align}$
That is, $-2-2\sqrt{3}i$.
Hence, ${{\left( 1-i\sqrt{3} \right)}^{2}}=-2-2\sqrt{3}i$.
