Equation of a straight line 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.
 A: 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.
A: 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.
