Answer
The two square roots are:
\[
-\frac{\sqrt{6}}{2}+\frac{\sqrt{2}}{2} i \quad \text { , } \quad \frac{\sqrt{6}}{2}-\frac{\sqrt{2}}{2} i
\]
Work Step by Step
\[
1-i \sqrt{3}
\]
The modulus of a complex number in the standard form $x+y i$ is:
\[
\begin{array}{c}
\sqrt{x^{2}+y^{2}} =r\\
\sqrt{1^{2}+(-\sqrt{3})^{2}}=r \\
\sqrt{3+1}=r \\
\sqrt{4}=r \\
2=r
\end{array}
\]
The $\theta$ angle is the least positive angle for which:
\[
\frac{1}{2}=\frac{x}{r}=\cos \theta
\]
,
\[
-\frac{\sqrt{3}}{2}=\frac{y}{r}=\sin \theta
\]
Thus,
\[
300^{\circ}=\theta
\]
Using the definition of a complex number in trigonometric form:
\[
\begin{aligned}
&=(\cos \theta+i \sin \theta)r \\
=\left(\cos 300^{\circ}+i \sin 300^{\circ}\right)2=z
\end{aligned}
\]
Using the roots theorem:
\[
r^{\frac{1}{n}}\left[\cos \left(\frac{\theta}{n}+\frac{360^{\circ}}{n} k\right)+i \sin \left(\frac{\theta}{n}+\frac{360^{\circ}}{n} k\right)\right]=w_{k}
\]
When $0=k$ :
\[
\begin{array}{c}
2 \frac{1}{2}\left[\cos \left(\frac{300^{\circ}}{2}+\frac{360^{\circ}}{2}(0)\right)+i \sin \left(\frac{300^{\circ}}{2}+\frac{360^{\circ}}{2}(0)\right)\right]=u_{0} \\
=\left[\cos 150^{\circ}+i \sin 150^{\circ}\right]\sqrt{2} \\
=\left[\frac{-\sqrt{3}}{2}+\frac{1}{2} i\right]\sqrt{2}
\end{array}
\]
Multiply:
\[
=-\frac{\sqrt{6}}{2}+\frac{\sqrt{2}}{2} i
\]
When $1=k$ :
\[
\begin{array}{c}
2^{\frac{1}{2}}\left[\cos \left(\frac{300^{\circ}}{2}+\frac{360^{\circ}}{2}(1)\right)+i \sin \left(\frac{300^{\circ}}{2}+\frac{360^{\circ}}{2}(1)\right)\right]=w_{1} \\
=\sqrt{2}\left[\cos \left(150^{\circ}+180^{\circ}\right)+i \sin (180+150)\right] \\
=\left[\cos 330^{\circ}+i \sin 330^{\circ}\right] \sqrt{2}\\
=\left[\frac{\sqrt{3}}{2}-\frac{1}{2}\right]\sqrt{2}
\end{array}
\]
Multiply:
\[
=\sqrt{\frac{\sqrt{6}}{2}-\frac{\sqrt{2}}{2} i}
\]
