How do you know what part of the semicircle it is when $\arg\dfrac{z-2}{z+2} = \dfrac\pi2$

Question

I'm not quite sure how to determine whether the semicircle is below or above the x-axis, because in $\arg\dfrac{z-2}{z+2} = \dfrac\pi2$, it lay above the x-axis with a locus of $(4-x^2)^{1/2}$.

Answer 1

$$\dfrac{x-2-iy}{x+2+iy}=\dfrac{(x-iy)^2-4}{(x+2)^2+y^2}=\dfrac{x^2-y^2-4-2xyi}{(x+2)^2+y^2}$$

Using atan2, $$x^2-y^2-4=0$$ and $$-2xy>0\iff xy<0$$