2

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.

mnmn1993
  • 435

1 Answers1

5

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.

Martin R
  • 113,040
  • 1
    Completely transparent. – Lubin Mar 06 '18 at 20:33
  • Thanks! May I know that why $T_1$ is the map from upper half-disk onto the second quadrant? If I express it as coordinate $T_1=\dfrac{r^2-1+2ir\sin\theta}{r^2+1+2r\cos\theta}=u+iv$. I can not find $r$ in term of $u$ and $v$. Or there is a simple method? – mnmn1993 Mar 07 '18 at 04:12
  • 1
    @mnmn1993: Use that Möbiustranformations map lines and circles onto lines and circles, and (as conformal maps of the extended plane) preserve angles and orientation. For example, $T_1(1)=0$, $T_1(0)=-1$, $T_1(-1)=\infty$ implies that the segment $[-1, 1]$ is mapped onto the negative real axis. Then show that the upper half of the unit circle is mapped to the positive imaginary axis. – Martin R Mar 07 '18 at 06:37
  • Sorry I still dont know why it is onto by your augment, it seems that you use the fact that the connected component is mapped to the connected component by Möbius map? – mnmn1993 Mar 07 '18 at 08:43
  • 1
    @mnmn1993: That is correct. For example, since $T_1$ maps the (extended) real axis onto itself, it maps the upper half-plane to the either the upper or the lower half-plane. Since $T_1(i) = i$ it must be the upper half-plane. One can also use an orientation argument: The upper half-plane lies to the right when moving from $1$ to $0$ on the real axis, therefore image lies to the right when moving from $T_1(1) = 0$ to $T_1(0) = -1$. – Martin R Mar 07 '18 at 08:50
  • Is this true for conformal map? – mnmn1993 Mar 07 '18 at 08:53
  • 1
    @mnmn1993: A conformal map is orientation-preserving. – Martin R Mar 07 '18 at 08:56
  • Where can I find a proof about the orientation principle? I can not find one in conway's book. Thanks – mnmn1993 Mar 07 '18 at 09:36
  • 1
    @mnmn1993: this might help as a starting point https://math.stackexchange.com/questions/687487/conformal-map-iff-holomorphic. The essential idea is that when viewing $f$ as a mapping of $\Bbb R^2$ into $\Bbb R^2$, the Jacobian matrix is a (scaled) rotation, because of the Cauchy-Riemann equations. – Martin R Mar 07 '18 at 09:42
  • How do you prove the second one? If i want to show $f(z)=z^{2}$ is a conformal map from the second quadrant to lower half plane, the first direction is easy. But for the inverse map, I will need to use $g(w)=w^{\frac{1}{2}}$, which let us get into the trouble of square root again. – JacobsonRadical Apr 05 '18 at 05:14
  • @JacobsonRadical: $f(z)=z^2$ is an injective, holomorphic function from the second quadrant onto the lower half-plane, therefore it has a (holomorphic) inverse which maps the lower half-plane onto the second quadrant. – For $w$ in the lower half-plane, $f^{-1}(w)$ is the (unique) square root of $w$ lying in the second quadrant. – Martin R Apr 05 '18 at 07:53
  • @JacobsonRadical Think in polar coordinates! – Martin R Apr 05 '18 at 14:38
  • @MartinR Oh I got you.. Sorry for the bad question. – JacobsonRadical Apr 05 '18 at 16:06