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I've been trying to prove the statement above for sometime and I feel like I am close. I was wondering if I could get some help with my ideas, to see if they are even close.

I have that $\arg \frac{z-1}{z+1} = \arg z-1 - \arg z+1$.

Also if we let $z = x + iy$, then clearly $z-1 = x-1 + iy$ and $z+1 = x+1 - iy$, from this we can see that $\operatorname{Im}z-1 = \operatorname{Im} z+1$.

Suppose $\arg z-1 = \beta$, $\arg z+1 = \theta$. I'm pretty sure the statement above implies that $\cos(\beta - \frac \pi 2) = \sin \theta$. And $\cos\theta_1 = \sin\theta_2$ for when $\theta_1 = \theta_2 = \frac{\pi}{4}$ so $\beta = \frac{3\pi}{4}$ and $\theta = \frac{\pi}{4}$ and this $\beta - \theta = \arg z-1 - \arg z+1 = \frac{\pi}{2}$.

If this is the correct logic at all I really need to fill in some holes, if not can I please have some hints at the right path to take.

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  • $\begingroup$ I can't recommend enough the use of parentheses. For example, I would normally read $\arg z - 1 - \arg z + 1$ to be equal to $0$. What I mean is if you think obvious that it should be $\arg (z - 1) - \arg (z + 1)$, then why not $\arg (z - 1 - \arg z) + 1$ or $\arg (z - 1 - \arg z + 1)$. And would you then write $(\arg z) - 1$ if you wanted to subtract $1$ from the argument of $z$? $\endgroup$ – Ennar Sep 10 '17 at 2:06
  • $\begingroup$ Sorry, that is just me being lazy. WIll keep in mind in future. $\endgroup$ – student_t Sep 10 '17 at 4:12
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$\displaystyle |z|=1\iff z=e^{i\theta}$

$\displaystyle w=\frac{z-1}{z+1}=\frac{e^{i\theta}-1}{e^{i\theta}+1}=\frac{e^\frac{i\theta}2}{e^\frac{i\theta}2}\left(\frac{e^\frac{i\theta}2-e^\frac{-i\theta}2}{e^\frac{i\theta}2+e^\frac{-i\theta}2}\right)=\frac{2i\sin(\frac{\theta}2)}{2\cos(\frac{\theta}2)}=i\tan(\frac{\theta}2)\quad$ is pure imaginary.

Also modulo $2\pi$ : $\quad\Im(z)>0\iff\theta\in]0,\pi[\iff\frac{\theta}2\in]0,\frac{\pi}2[\iff\tan(\frac{\theta}2)>0\iff \arg(w)=+\frac{\pi}2$.

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$$w=\dfrac{z-1}{z+1}=\dfrac{z-1}{z+1}\dfrac{\overline{z}+1}{\overline{z}+1}=i\dfrac{2{\bf Im\,}z}{|z+1|^2}$$ which shows $${\bf Re\,}w=0~~~,~~~{\bf Im\,}w=\dfrac{2{\bf Im\,}z}{|z+1|^2}>0$$ thus $\color{blue}{\arg w = \dfrac{\pi}{2}}$.

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  • $\begingroup$ Wow, I was overcomplicating it entirely. It makes sense to show that it must be pure imaginary then the arg has to be ${\pi}{2}$ depending on the sign! Thank you so much, all these answers were very insightful! $\endgroup$ – student_t Sep 10 '17 at 4:13
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Hint:   let $w = \frac{z-1}{z+1}\,$, then:

$$\require{cancel} w + \bar w = \frac{z-1}{z+1}+\frac{\bar z-1}{\bar z+1} = \frac{z \bar z + \bcancel{z} - \cancel{\bar z} - 1 \;+\; z\bar z - \bcancel{z} + \cancel{\bar z} - 1}{|z+1|^2}=2\,\frac{|z|^2-1}{|z+1|^2} $$

It follows that $|z|=1 \implies w + \bar w = 0 = 2 \operatorname{Re}(w)\,$, so $\operatorname{Arg} w = \pm \frac{\pi}{2}\,$. All that remains to be shown is that $\operatorname{Im}(z) \gt 0 \implies \operatorname{Im}(w) \gt 0\,$, which follows by similarly calculating $2i \operatorname{Im}(w) = w - \bar w\,$.

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