$\displaystyle Arg\left (\frac{i\omega}{2-i\omega} \right )=Arg(i\omega)-Arg(2-i\omega)$

Solution:

$\displaystyle \pi-arctan\left (\frac{\omega}{2} \right )$

$\displaystyle \omega>0$

My attempt:

$\displaystyle Arg(i\omega)=\frac{\pi}{2}$

$\displaystyle Arg(2-i\omega)=-arctan\left (\frac{\omega}{2} \right )$

$\displaystyle Arg(i\omega)-Arg(2-i\omega)=\frac{\pi}{2}-\left (-arctan\left (\frac{\omega}{2} \right ) \right )=\frac{\pi}{2}+arctan\left (\frac{\omega}{2} \right )$

Can somebody explain why the solution differs from my solution?

I have tested Arg(iw) with any number w>0 and they all give me pi/2?

Also I don't understand how they got the subtraction when mine ends up being a sum?