Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am trying to find a conformal map of slit unit disk (slit in negative real axis) i.e $${{z: |z|<1, z\notin (-1,0]}}$$ on to the unit disk that takes $\sqrt2 -1 $ to $0$.

This is what I think,

I can see $\sqrt{z} $ taking the slit disk to disk ( half disk) in the right half plane, and then rotating counterclockwise gives the disk in the upper half plane and the mapping $(\frac {1-z}{1+z} )^2$ maps to the upper half plane and the map $\frac {z-i}{z+i}$ maps to open unit disk.

I tried to compose and get the image of $\sqrt2 -1 $ under this composition. I am getting some frustrating crap.

If I knew that this $\sqrt2 -1 $ gets mapped to some $\alpha$ then I would compose the above function (composition function) with the map $(\frac {z-\alpha}{1-\bar\alpha z} )$ to get the image of $\sqrt2 -1 $ as $0$.

I also think this map is not unique.

So the question is if my work correct? If not what am I doing wrong? Can someone give me the explicit formula for this. Thanks in advance.

share|cite|improve this question
$\alpha:=\sqrt{2}-1$ is on the symmetry axis of your slit disk. In choosing the intermediate steps you should take care that this symmetry doesn't get lost. In the end the point $\alpha$ will be mapped onto some point $\beta>0$. – Christian Blatter Dec 25 '12 at 9:24
Of course your map is not unique: You can rotate the disk at the end. – Hagen von Eitzen Dec 25 '12 at 9:42
@ChristianBlatter, Can you show me more rigorously what symmetry you are talking about please! – Deepak Dec 28 '12 at 16:38
up vote 1 down vote accepted

We have a sequence of maps $$f_i:\ D_{i-1}\to D_i,\quad z_{i-1}\to z_i\qquad (1\leq i\leq 5)\ .$$ Here $z_i$ does not denote a certain point in the $z$-plane, but the coordinate variable in the $i$th auxiliary complex plane. $Z_0:=\sqrt{2}-1$ is the $z_0$-coordinate of a certain point $Z$ we are interested in.

$D_0$ is the unit disk in the $z_0$-plane minus the points $z_0\leq0$. The map $$f_1:\ z_0\mapsto z_1:={\rm pv}\sqrt{z_0}$$ maps $D_0$ onto the right half $D_1$ of the unit disk in the $z_1$-plane. Thereby the point $Z_0$ is mapped onto a point $Z_1\in\ ]0,1[\ $.

The Moebius map $$f_2:\ z_1\mapsto z_2:=-{z_1-i\over z_1+i}$$ maps $i$ to $0$ and $-i$ to $\infty$. Furthermore $f_2(0)=1$, $\ f_2(1)=i$. From general properties of Moebius maps it then follows that $D_2:=f_2(D_1)$ is the first quadrant, and that $f_2$ maps the real axis onto the unit circle. Therefore $Z_2=f_2(Z_1)$ is a point between $1$ and $i$ on the unit circle.

The map $$f_3:\ z_2\mapsto z_3:=z_2^2$$ maps the first quadrant $D_2$ onto the upper half-plane $D_3$, whereby $f_3(1)=1$, $\ f_3(i)=-1$, and the quarter unit circle in $D_2$ is mapped onto the upper half of the unit circle in $D_3$. Therefore the point $Z_3:=f_3(Z_2)$ is lying on this upper half of the unit circle, too.

The Moebius map $$f_4:\ z_3\mapsto z_4:=i{z_3-i\over z_3+i}$$ maps the upper half plane $D_3$ onto the unit circle $D_4$. Thereby $f_4(-1)=-1$, $\ f_4(1)=1$, and the unit circle of the $z_3$-plane is mapped onto the real axis of the $z_4$-plane. It follows that $Z_4=f_4(Z_3)$ is a real number between $-1$ and $1$.

Doing the calculations $Z_4$ should simplify to an expression defining a real number $\alpha\in\ ]{-1},1[\ $. Letting $$f_5:\ z_4\mapsto{z_4-\alpha\over 1-\alpha z_4}$$ you finally arrive at the required map $$f:=f_5\circ f_4\circ f_3\circ f_2\circ f_1\ .$$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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