2
$\begingroup$

Finding a Mobius transformation that maps the upper half-plane $\{Im(z)>0\};z \in \mathbb{C}$ to the inside of a unit-disk, such that point $i$ is mapped to $0$ and $\infty$ to $-1$.

Okay, to be clear, I know that a Mobius transformation $w$ is of the form:

$$w=\frac{az+b}{cz+d};ad-bc\neq0.$$ I am very aware of what a unit disk is. I have done assignments like finding an mobius transformation that maps some points ($\mathbb{C}$) $a,b,c$ to $d,e,f$. -Where these points were not infinity.

But nothing like this problem that I have here? How is this done? I think I just need one more point and know it's picture to be able to figure is out. But how?

$\endgroup$
4
$\begingroup$

$i$ is mapped to $0$, i.e. $\dfrac{ai+b}{ci+d} = 0$. So $b=-ai$.

$\infty$ is mapped to $-1$, i.e. $\dfrac{a}{c}=-1$. So $c=-a$.

We can choose $a$ to be equal to $1$. So $b=-i$ and $c=-1$, giving for some $d$ $$w(z)=\dfrac{z-i}{-z+d}$$

We want to choose $d$ so that $w$ maps the upper half-plane to the interior of the unit disk. $|z-i|$ is the distance from $z$ to $i$; and the upper half-plane is the set of points that are closer to $i$ than to $-i$. So if we choose $d=-i$, we will have $0 \le |w(z)| < 1$ for $z$ in the upper half-plane, as required.

$\endgroup$
  • $\begingroup$ This is great. I thank you very much for the insight. All the best. $\endgroup$ – Bozo Vulicevic Jan 6 '17 at 19:50
2
$\begingroup$

One very useful Mobius transformation to remember is the Cayley transform defined by $$ f(z)=\frac{z-i}{z+i}$$

This maps the upper half plane to the open unit disk because the upper half plane is precisely the set of points in $\mathbb{C}$ which are closer to $i$ than to $-i$. Note also that $f(i)=0$.

This map doesn't quite fit your requirements because $f(\infty)=1$, but $-f(z)$ works instead.

$\endgroup$
1
$\begingroup$

When wanna map three points $\{z_1,z_2,z_3\}$ to $\{w_1,w_2,w_3\}$, use the formula $$\frac{w-w_1}{w-w_3}\frac{w_2-w_3}{w_2-w_1}=\frac{z-z_1}{z-z_3}\frac{z_2-z_3}{z_2-z_1}$$ In this practice we have two points and need an extra point, so we choose $0$ to $1$, then we map $\{i,\infty,0\}$ to $\{0,-1,1\}$. If one of the points were the point at $\infty$ (here $z_2=\infty$) we delete it from formula: $$\frac{w-w_1}{w-w_3}\frac{w_2-w_3}{w_2-w_1}=\frac{z-z_1}{z-z_3}$$ $$\frac{w-0}{w-1}\times\frac{-1-1}{-1-0}=\frac{z-i}{z-0}$$ after siplifying $$w=\frac{i-z}{z+i}$$

$\endgroup$
  • $\begingroup$ You can't choose the third mapping arbitrarily, because $w$ has to map the upper half-plane to the interior of the unit disk. Your choice of $0$ to $1$ works here, but it looks like blind luck! $\endgroup$ – TonyK Jan 6 '17 at 19:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.