[Math] Why the map $ w=-1/2(z+1/z)$ maps upper half disk to upper half plane

Question

I have read some posts about this map( 1 and 2 ). But I still do understand about it.

If I express as coordinate, if $z=re^{i\theta}=x+iy$, then $$f=w=u+iv=\dfrac{-1}{2}(z+\dfrac{1}{z})=\dfrac{-1}{2}(\cos\theta(r+\dfrac{1}{r})+i\sin\theta(r-\dfrac{1}{r}) )$$
$$f=\dfrac{-1}{2}(\dfrac{x(x^2+y^2+1)}{x^2+y^2})+i\dfrac{y(x^2+y^2-1)}{x^2+y^2})$$
I can see that $f$ maps the upper half disk to subset of upper half plane, but I can not find a representation of $x$ and $y$ in term of $u$ and $v$ to show that the map is one to one and onto from upper half disk to upper half plane.

That is , how to show that for any $u\in \mathbb{R}$ and $v\in \mathbb{R}^+$, there are $x$ and $y$ from the upper half circle.

On the other hand, if I write $z^2+2wz+1=0$, I can solve that $z=-w\pm\sqrt{w^2-1}$ with $\sqrt{w^2-1}=|w^2-1|^{1/2}e^{i Arg(w^2-1)}$. Denote them by $z^+$ and
$z^-$. Suppose I know that $z^+$ is the desired result root.

But how can I show that it is a bijective map from upper half plane to upper half disk? It seems that it is hard to compute the argument of $z=-w\pm\sqrt{w^2-1}$ and its norm for $w\in \mathbb{H}$.

In the post 2, it was stated that if we know $z^+(i)=-i+\sqrt{2}i\in\mathbb{H}$ then it is enough to say $z^+$ is the required map, but why?

And in general, why $z^+$ is the required map? It was stated that in 1 and 2 that $z^+z^-=1$, so if one of them is outside the circle, then another one must be inside the circle, but I think it can be that sometimes $z^+ $ may be inside the circle and sometimes $z^+ $ may be outside the circle, why there must be a fixed one?

And finally why the branch is taken $(-1,1)$ since I think the negative real axis is enough. Sorry for so many questions.

Answer 1

There is an alternative approach, which avoids the hassle with (holomorphic branches of) square roots: We have $$ f(z) = -\frac 12 (z + \frac 1z) = \frac{\left( \frac{z-1}{z+1}\right)^2 + 1}{\left( \frac{z-1}{z+1}\right)^2 - 1} = T_2(S(T_1(z)) $$ where:

• $T_1(z) = \frac{z-1}{z+1}$ is a Möbiustransformation mapping the upper half-disk onto the second quadrant,
• $S(z) = z^2$ maps the second quadrant bijectively onto the lower half-plane, and
• $T_2(z) = \frac{z+1}{z-1}$ is a Möbiustransformation mapping the lower half-plane onto the upper half-plane,

so that the composition is a one-to-one mapping from the upper half-disk to the upper half-plane.