Precalculus (6th Edition) Blitzer

$-2-2\sqrt{3}i$
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, ${{\left( 1-i\sqrt{3} \right)}^{2}}=-2-2\sqrt{3}i$.