Finding a Conformal Function for Mapping D1 onto the Unit Disc D2

[skrat]

Homework Statement

Find conformal function that maps $D_1=\left \{ z;Re(z)>0,0<Im(z)<2\pi \right \}$ on to $D_2$ where $D_2$ is unit disc.

Homework Equations

The Attempt at a Solution

Ok, I haven't got any problems with conformal mapping but I have huge problems with function composition and I would need some help here...

Firstly, $f_1=e^z$ maps from $D_1$ into unit disc without $Re(z)>0$ axis, so to this object I now apply $f_2=\sqrt{z}$ which gives me upper half of unit disc (real exis not included!).

Now $f_3=\frac{z+1}{1-z}$ maps the upper unit disk into first quadrant. Applying $f_4=z^2$ extends my area on to upper half plane. Rotating it with $f_5=ze^{-i\pi /2}$ and again using Mobius transformation $f_6=\frac{i-z}{z+i}$ gives me that unit disk.

Now the question is of course what $f=f_6\circ f_5\circ f_4\circ f_3\circ f_2\circ f_1$?

Does it even make any sense to do that?

 

[Fredrik]

Homework Statement

Find conformal function that maps $D_1=\left \{ z;Re(z)>0,0<Im(z)<2\pi \right \}$ on to $D_2$ where $D_2$ is unit disc.

Homework Equations

The Attempt at a Solution

Ok, I haven't got any problems with conformal mapping but I have huge problems with function composition and I would need some help here...

Firstly, $f_1=e^z$ maps from $D_1$ into unit disc without $Re(z)>0$ axis,

$f_1(1+i\pi)=e^{1+i\pi}=e^1 e^{i\pi}= -e$ is not in the unit disc.

 

[skrat]

Of course not. It even doesn't have to be.

There is a typo in my original post. Instead of $D_1=\left \{ z;Re(z)>0,0<Im(z)<2\pi \right \}$ it is $D_1=\left \{ z;Re(z)<0,0<Im(z)<2\pi \right \}$