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What is the locus given by the following equation?

$$arg\left(\frac{z+1}{z-1}\right)=\frac{\pi}{2}$$

I know that $$arg\left(\frac{z_1}{z_2}\right)=arg(z_1)-arg(z_2)$$

and that the loci must emanate from -1 and +1 on the real axis. The $\pi/2$ also suggest to me that there is a $90^\circ$ angle at work, but I don't know how to proceed from there. So a more substantive explanation would be most appreciated!

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It's pure high-school geometry: the upper semi-circle with diameter $[-1,1]$.

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  • $\begingroup$ Thanks, yes you are right. It is high school mathematics. That's the level I am at. $\endgroup$ – Yoon Mar 27 '17 at 2:31
  • $\begingroup$ @Bernard It should be the lower semi-circle, rather. $\endgroup$ – dxiv Mar 27 '17 at 2:33
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Algebraic alternative: first note that $\arg w = \frac{\pi}{2}$ iff $\operatorname{Re} w = 0$ and $\operatorname{Im} w \gt 0\,$. Then:

  • $\require{cancel}0 = 2 \, \operatorname{Re} \cfrac{z+1}{z-1} = \cfrac{z+1}{z-1}+\cfrac{\bar z+1}{\bar z-1}=\cfrac{z \bar z - \cancel{z} + \bcancel{\bar z} - 1 + z \bar z - \bcancel{\bar z} + \cancel{z} - 1}{|z-1|^2}=\cfrac{2(|z|^2 - 1)}{|z-1|^2} \implies \bbox[3px, border:1px solid black]{|z|^2 = 1}\,$

  • $0 \lt 2 \,\operatorname{Im} \cfrac{z+1}{z-1} = \cfrac{1}{i}\left(\cfrac{z+1}{z-1}-\cfrac{\bar z+1}{\bar z-1}\right)=\cfrac{1}{i} \, \cfrac{-2(z-\bar z)}{|z-1|^2}=- \cfrac{4 \operatorname{Im} z}{|z-1|^2}$ $\implies$ $\bbox[3px, border:1px solid black]{\operatorname{Im} z \lt 0}$

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