Argument and modules of complex numbers Can you help me to find the argument and modules of this complex number please? 
$$z=\frac{2-i}{-\sqrt {3}-2i}$$
 A: Hint. Note that
$$z=\frac{2-i}{-\sqrt {3}-2i}\cdot\frac{-\sqrt {3}+2i}{-\sqrt {3}+2i}=\frac{(2-i)(-\sqrt {3}+2i)}{(-\sqrt {3})^2+2^2}.$$
After putting $z$ in the cartesian form $x+iy$, then $|z|=\sqrt{x^2+y^2}$. In order to find  the argument, see wiki-page ($z$ is in the second quadrant!). It is not a "nice" angle... It would be nice with the numerator equal to $2-2i$ or $1-i$.
A: $$z=\frac{2-i}{-\sqrt {3}-2i}\cdot\frac{-\sqrt {3}+2i}{-\sqrt {3}+2i}=\frac{-2\sqrt3+2i-i\sqrt3-2i^2}{3+4}=$$
$$=\frac{2-2\sqrt3+(2-\sqrt3)i}{7}=\frac{2-2\sqrt3}{7}+\frac{2-\sqrt3}{7}i=a+bi$$
then
$$|z|=\sqrt{a^2+b^2}$$
$$\arg z=\arctan\frac{a}{b}$$
A: Argument of $z$,
\begin{align}
\arg(z)&=\arg\left(\dfrac{2-i}{-\sqrt{3}-2i}\right)\\
&=\arg(2-i)-\arg(\sqrt{3}-2i)\\
&=\left[-\tan^{-1}\left(\dfrac{1}{2}\right)\right]-\left[-\pi+\tan^{-1}\left(\dfrac{2}{\sqrt3}\right)\right]\\
&=\pi-\left[\tan^{-1}\left(\dfrac{1}{2}\right)+\tan^{-1}\left(\dfrac{2}{\sqrt3}\right)\right]\\
&=\pi-\tan^{-1}\left(\dfrac{\frac{1}{2}+\frac{2}{\sqrt3}}{1-\frac{1}{2}\cdot\frac{2}{\sqrt3}}\right)\\
&=\pi-\tan^{-1}\left(\dfrac{4+\sqrt3}{2\sqrt3-2}\right)
\end{align}
And modulus of $z$,
\begin{align}
|z|&=\left|\dfrac{2-i}{-\sqrt{3}-2i}\right|\\
&=\dfrac{|2-i|}{|-\sqrt{3}-2i|}\\
&=\dfrac{\sqrt{2^2+1^2}}{\sqrt{(\sqrt{3})^2+2^2}}\\
&=\dfrac{\sqrt{35}}{7}
\end{align}
