Calculating absolute value and argument of a complex number I want to calculate the absolute value and argument of the complex number $a = \left(\sqrt{3} - i\right)^{-2}$. 
In order to calculate these two values I tried to reform the number into the form $z = x + y \cdot i$:
$$a =\left(\sqrt{3} - i\right)^{-2} = \frac{1}{\left(\sqrt{3}-i\right)^2}  = \frac{1}{2 +\left(-2\sqrt{3}\right)}$$
$$ = -\frac{2}{8} - i \frac{-2\sqrt{3}i}{4-12} = -\frac{1}{4} - i \frac{\sqrt{3}i}{4}$$
I now want to calculate the polar form of a:
$$r = |z| = \sqrt{x^2 + y^2} = \sqrt{\frac{1}{8} + \left( -1 -2i \left( \frac{\sqrt{3}i}{4}\right) + \left(\frac{\sqrt{3}i}{4}\right)^2 \right)}$$
$$ = \sqrt{\frac{1}{8} + \left( -1 -2i \left( \frac{\sqrt{3}i}{4}\right) + \frac{3 + 2 \sqrt{3}i - 1}{16} \right)} =
 \sqrt{\frac{1}{8} + \left( -1 -i \left( \frac{\sqrt{3}i}{2}\right) + \frac{3 + 2 \sqrt{3}i - 1}{16} \right)} $$
$$=\sqrt{\frac{1}{8} + -1 \frac{\sqrt{3}-1}{2} + \frac{2 + 2 \sqrt{3}i}{16}}$$
but finally am stuck here. Can you please help me to go on? Is there a better way to get to the absolute value and argument?

Rewrite:
\begin{align}
(\sqrt{3}-i)^{-2}&=\frac{1}{(\sqrt{3}-i)^{2}}\\
&=\frac{1}{2-2\sqrt{3}\ i}\\
&=\frac{1}{2-2\sqrt{3}\ i}\cdot\frac{2+2\sqrt{3}\ i}{2+2\sqrt{3}\ i}\\
&=\frac{2+2\sqrt{3}\ i}{2^2-(2\sqrt{3}\ i)^2}\\
&=\frac{2+2\sqrt{3}\ i}{16}\\
&=\frac18+\frac18\sqrt{3}\ i
\end{align}
It should be easy from this to obtain its absolute value and argument. The rest, I leave it to you. Good luck! :)

It looks like you didn't know to evaluate the absolute value and argument of $|z|$. The absolute value can be obtained by using this formula
$$
|z|=\sqrt{x^2+y^2}
$$
and the argument of $z$ is
$$
\arg(z)=\arctan\left(\frac yx\right).
$$
In this case, we have $x=\Re(z)=\frac18$ and $y=\Im(z)=\frac18\sqrt{3}$. You may refer to this link to learn more about complex number. I hope this helps.
