<s>[INST] 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?
 [/INST] I would start by finding the polar form of $z = \sqrt{3} - i$. This corresponds to a nice triangle, the so called 30-60-90 triangle. Here one leg is of length $\sqrt{3}$ and the other is of length 1 (in the negative imaginary direction). The hypotenuse of this triangle is the magnitude of $z$, so $|z|=2$.
The argument of this $z$ is $\theta = -\pi/6$.
Now the number you want is $$a = 1/z^2 = \frac{1}{|z|^2e^{i2\theta}} = \frac{1}{|z|^2} e^{-2i\theta}.$$
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