I am studying Complex Analysis on my own and am having a bit of difficulty with finding the locus of $0<\arg(\frac{z+i}{z-i})<\pi/4$ rigorously. We can see geometrically (using Inscribed Angle Property) that the locus is the part of the plane that's to the right of the y-axis, minus a certain circle on the center (I calculated this circle to be $(x-1)^2+y^2=2$ specifically), but can this be calculated without this sort of geometrical intuition?
ATTEMPT: I tried plugging $z=x+iy$. We get $\Re(z)=\frac{x^2+y^2-1}{(x^2+(y-1)^2}$ and $\Im(z)=\frac{2x}{(x^2+(y-1)^2}$
So my thought was to try and compare these two to $r\cos(t)$ and $r\sin(t)$ respectively, and then maybe use the formula $\sin^2(t)+\cos^2(t)=1$, but I didn't get very far.