Equation of a straight line

Question

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.

Answer 1

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.

Answer 2

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.