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Why is the image of $g(w)=\log\left({1+w\over 1-w}\right)$ for $0<\arg(w)<\pi$ all lie on a straight line?

Sorry about the confusing statement in the beginning.

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    Possibly related: http://math.stackexchange.com/questions/118868/for-complex-z-z-1-implies-textre-left-frac1-z1z-right-02012-03-22

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By the open mapping theorem, since $\log({1 + w \over 1 - w})$ is analytic, the image of any open set under $\log({1 + w \over 1 - w})$ will be open and therefore not be on a line.

So I'll guess you mean just the image of the $0 < arg(w) < \pi$ portion of unit circle $|w|= 1$ under this map. In that case, write $w = e^{i\theta}$. Then ${1 + w \over 1 - w} = {1 + e^{i\theta} \over 1 - e^{i\theta}}$ $ = {e^{i\theta \over 2} + e^{-{i\theta \over 2}} \over {e^{i\theta \over 2} - e^{-{i\theta \over 2}}}} $ $= -i\cot({\theta\over 2})$ Taking logs of this you get $\ln(\cot({\theta \over 2}))-i{\pi \over 2}$. This is on the line $y = -i{\pi \over 2}$ in the complex plane.

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My notation is a bit different from yours, and you would do well to check this for mistakes, but here's what I have so far. We are taking the complex logarithm of a linear fractional or Möbius transformation, so there is probably a good geometric-complex analytic perspective on our function.

If $w=g(z)=\log\left(\frac{1+z}{1-z}\right)$ and $z=re^{i\theta}$, so that $\theta=\arg z$, then $ e^w=\frac{1+z}{1-z}\quad\implies\quad z=\frac{e^w-1}{e^w+1}=\sinh\tfrac{w}{2}. $

If we treat $w=u+iv$ as a function from $(r,\theta)\in R$ to $(u,v)\in R$ for $R=[0,\infty)\times(-\pi,\pi]\subset\mathbb{R}^2$, then $ e^u(\cos v+i\sin v)= e^w= \frac{ (1-r^2)+i(2r\sin\theta) }{ (1+r^2)-(2r\cos\theta) }. $ Note that the denominator is always nonnegative since it equals $(1\mp r)^2$ for $\cos\theta=\pm1$. Furthermore, $ e^u=|e^w| =\frac{\sqrt{(1-r^2)^2+(2r\sin\theta)^2}}{1+r^2-2r\cos\theta} =\sqrt{\frac{1+r^2+2r\cos\theta}{1+r^2-2r\cos\theta}} $ and $ \tan v=\frac{2r\sin\theta}{1-r^2}, \qquad \sin v=\frac{2r\sin\theta}{R} \quad \& \quad \cos v=\frac{1-r^2}{R} $ for $ R^2 =(1+r^2)^2-(2r\cos\theta)^2 =(1-r^2)^2+(2r\sin\theta)^2 =1+r^4-4r^2\cos2\theta. $ If we are expecting the image of $g$ to be a straight line for $\theta\in(0,\pi)$, that would mean that some linear combination of $u$ and $v$ is constant, or that $\frac{\partial u}{\partial \theta}$ and $\frac{\partial v}{\partial \theta}$ are proportional, or that $\frac{du}{dv}$ (or its reciprocal) is constant. I got $ \frac{\partial u}{\partial \theta}= \frac{4r^2\sin\theta\cos\theta} {(1+r^2)^2-(2r\cos\theta)^2} $ and $ \frac{\partial v}{\partial \theta}= \frac{2r(1-r^2)\cos\theta} {(1+r^2)^2-(2r\cos\theta)^2} $ or $ \frac{du}{dv}=\frac{2r}{1-r^2}\sin\theta=\tan v, $ which would give $ u=\int du=\int\tan v\,dv=-\ln|\cos v|+c_1 \quad\implies\quad e^u \cos v=\text{constant} $ and contradicts what we already have above, namely $ e^u \cos v = \frac{ 1-r^2 }{ 1+r^2-2r\cos\theta } $ which exhibits a dependence on $\theta$. So there must be an error in this somewhere (can anyone find it?).

However, when $r=1$, we have $\frac{dv}{du}=0$, which is a horizontal line (or a symmetric square wave on $(-\pi,\pi)$ with dirac/delta derivative) $v=\pm\frac{\pi}{2}$ (with the same sign as $\theta$).

I have checked some of this symbolically and graphically in a sage workbook, published here.

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    @Gary: Please nevertheless leave the question in a meaningful state compatible with the answer provided. Remember that the site isn't only for those who ask the questions but also for those who may find the questions and answers later.2012-02-18