# How to find transformation from the upper half plane into the right half plane?

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.

Any hint?

• How about you just rotate the plane?
– quid
Aug 5 '18 at 0:21
• @quid Ok, Am I going to use this transformation $T(z)=e^{i\theta_0}z$ then?.
– user486983
Aug 5 '18 at 0:28
• Yes. And what is $\theta_0$? Aug 5 '18 at 0:37
• @DavidG.Stork it's the angle of rotation, the principal argument
– user486983
Aug 5 '18 at 0:41
• 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 $c=0.$ 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.

• 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 $a d - b c > 0$. Aug 5 '18 at 20:19
• @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. Aug 5 '18 at 20:32
• Hi, I didn't know the book have answers.. I checked the answer and it's $w=-i(az+b)$
– user486983
Aug 7 '18 at 2:37
• Do you know how was constructed?
– user486983
Aug 7 '18 at 2:38
• Well, they're talking about linear transformation (actually affine transformations) not Möbius transformations. This is just the $c=0$ case of the above. Aug 7 '18 at 2:44