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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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I don't think your statement is true the way it is stated. If you take $\omega_\alpha=\alpha i$ (so $\arg(\omega)=\pi/2$), then $g(\omega)=i\arctan(2\alpha/(1-\alpha^2))$, which does not form a straight line. – Martin Argerami Feb 19 '12 at 0:03
Note that the question has been changed after an answer was given. See also my comments under bgins' answer on problems in the formulation of the question. – joriki Feb 19 '12 at 8:50
I think the image is only on a straight line when we also assume that $w$ has unit modulus. See my (mostly) corrected post for details. – bgins Feb 19 '12 at 16:18
Possibly related:… – Aryabhata Mar 22 '12 at 0:15

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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Why is $z=\mathrm e^{\mathrm i\theta}$? – joriki Feb 18 '12 at 16:40
@joriki: That's because if we let $z=e^{i\theta}$, we get the expression on the rhs being $\log((1+z)/(1-z))$ What I am asking can probably be reduced to why all the image points of the set of points $z=e^{i\theta}$ where $\theta\in (0,\pi)$ all lie on a straight line? – Gary Feb 18 '12 at 16:45
You're right. Corrected. Well, maybe we should try to demonstrate that the real and imaginary parts of $w$ lie on a straight line in the real plane, since thez're already parametrized by $\theta$. Why not try differentiating these with respect to $\theta$? – bgins Feb 18 '12 at 16:45
@Gary: If that's what you were trying to ask, you should modify the question accordingly. Currently I don't see anything in the question that implies $z=\mathrm e^{\mathrm i\theta}$. As far as I can see, the fact that we get the expression on the right-hand side to take a particular form if we assume $z=\mathrm e^{\mathrm i\theta}$ doesn't imply anything. – joriki Feb 18 '12 at 16:53
@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. – joriki Feb 18 '12 at 18:03

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