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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.

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$$\frac{z+i}{z-i} = \frac{(z+i)(\bar{z}+i)}{|z-i|^2} \Rightarrow \arg(\frac{z+i}{z-i}) = \arg((z+i)(\bar{z}+i))$$

That being said, note that the argument of a complex number of the form $Z=X+iY$ is in $(0,\frac{\pi}{4})$ if and only if $X>Y>0$. As $$(z+i)(\bar{z}+i) = x^2+y^2-1+2ix,$$ the desired condition is verified if and only if $$x^2+y^2-1 > 2x > 0,$$ which can be rewritten as $$x>0 \text{ and } (x-1)^2+y^2>2.$$

Thus the locus is the half-plane $\mathbb{R}_+\times\mathbb{R}$ minus the circle of radius $\sqrt{2}$ centered in $(1,0)$.

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    Brilliant, thanks!2012-11-08