Find the general form of the linear transformation which transforms

the upper half plane into the right half plane.

In my notes I have a Mobius transformation from the upper half plane to the unit circle $T(z)=e^{i\theta_0}\frac{z-z_0}{z-\overline z_0}$.

Also another transformation from the unit circle to the upper half plane $T(z)=(1-i)\frac{z-i}{z-1}$.

But I do not know how to construct the possible composition transformation from upper half plane into the right half plane.

Could someone help please?

Any hint?

  • 6
    $\begingroup$ How about you just rotate the plane? $\endgroup$
    – quid
    Aug 5 '18 at 0:21
  • $\begingroup$ @quid Ok, Am I going to use this transformation $T(z)=e^{i\theta_0}z$ then?. $\endgroup$
    – user486983
    Aug 5 '18 at 0:28
  • $\begingroup$ Yes. And what is $\theta_0$? $\endgroup$ Aug 5 '18 at 0:37
  • $\begingroup$ @DavidG.Stork it's the angle of rotation, the principal argument $\endgroup$
    – user486983
    Aug 5 '18 at 0:41
  • 1
    $\begingroup$ I'd be glad to tell you if I knew how. A Möbius transformation $T$ is of the form $$T(z) ={az+b\over cz+d}$$ where $ad-bc\neq 0,$ so the question apparently asks for conditions on $a,b,c,d.$ It looks easy if $c=0$ -- just a straightforward elaboration of the answers given in the comments -- but I haven't been able to either handle the $c\neq0$ case or to prove that $c=0.$ $\endgroup$
    – saulspatz
    Aug 5 '18 at 13:54

Simple, rotating 90 degrees to the right:

$$ T(z) = -iz $$


I think I've got it. The transformation $T$ must take the boundary of the upper half-plane to the boundary of the right half-plane. That is, it must take the real axis to the imaginary axis. If $$T(z)={az+b\over cz +d},\ ad-bc\neq=0,\tag{1}$$ we have that $x\in\mathbf{R}$ implies $$\Re\frac{(ax+b)\overline{(cx+d)}}{|cx+d|^2}=0\implies\Re(ax+b)\overline{(cx+d)}=0$$

Now for given, $a,b,c,d\in\mathbf{C},\ \Re(ax+b)\overline{(cx+d)}$ is a quadratic in $x$ that vanishes everywhere, so all the coefficients must be $0$. That is, $$\Re(a\overline{c})=\Re(a\overline{d}+b\overline{c})=\Re(b\overline{d})=0\tag{2}$$

Let us assume that $c\neq0.$ Then we may divide numerator and denominator in $(1)$ by $c$, or what is the same thing, we may assume that $c=1,$ so $(2)$ becomes $$\Re(a)=\Re(a\overline{d}+b)=\Re(b\overline{d})=0\tag{3}$$

From $c=1$ we have $T(-d)=\infty,$ but $\infty$ is on the imaginary axis so $d\in\mathbf{R},$ and from $(3),$ we have $\Re a=0$ and $\Re(ad+b)=0,$ so that $\Re b = 0.$ That is, $$T(z) = i\frac{\alpha z+\beta}{z+d}, \text { where } \alpha,\beta,d\in\mathbf{R}, \alpha d -\beta\neq0$$ which can obviously be re-written more symmetrically as $$T(z) = i\frac{\alpha z+\beta}{\gamma z+ \delta}, \text { where } \alpha,\beta,\gamma \delta\in\mathbf{R}, \alpha\delta-\beta\gamma\neq=0$$

However, this leaves open the possibility that $T$ maps the upper half-plane to the $left$ half-plane.

This leaves you with two things to do. First, finish off the $c\neq0$ case. (Hint: $\Re T(i)>0$.) Second, do the (easier) $c=0$ case.

I feel that there must be an easier way of seeing this, but I've not been able to find one.

  • 1
    $\begingroup$ Might be easier to start with the mapping of the upper half-plane to itself and then multiply by $-i$. From the formula for a linear-fractional transform mapping $z_1, z_2, z_3$ to $w_1, w_2, w_3$, we obtain that mappings of the real line to itself are given by $(a z + b)/(c z + d)$ with $a, b, c, d$ real. The condition that the direction is preserved is given by $T'(z) > 0$ for real $z$, or $a d - b c > 0$. $\endgroup$
    – Maxim
    Aug 5 '18 at 20:19
  • $\begingroup$ @Maxim T I don't have time to consider this closely just now, but on a quick read, it sounds goo. I'll think about it later. Thanks. $\endgroup$
    – saulspatz
    Aug 5 '18 at 20:32
  • $\begingroup$ Hi, I didn't know the book have answers.. I checked the answer and it's $w=-i(az+b)$ $\endgroup$
    – user486983
    Aug 7 '18 at 2:37
  • $\begingroup$ Do you know how was constructed? $\endgroup$
    – user486983
    Aug 7 '18 at 2:38
  • $\begingroup$ Well, they're talking about linear transformation (actually affine transformations) not Möbius transformations. This is just the $c=0$ case of the above. $\endgroup$
    – saulspatz
    Aug 7 '18 at 2:44

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