# Finding a conformal map of slit disk on to unit disk

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.

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$\alpha:=\sqrt{2}-1$ 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

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 $$f:=f_5\circ f_4\circ f_3\circ f_2\circ f_1\ .$$

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