How to find an analytic $f:G\to\mathbb C$, where $G=\{z:Re(z)>0\}$, such that $f(G)=D=\{z:|z|<1\}$ using a Mobius Transformation?

Question

The following example is taken from Conway's book, Functions of One Complex Variable I.

Find an analytic function $f:G\to\mathbb C$, where $G=\{z:Re(z)>0\}$, such that $f(G)=D=\{z:|z|<1\}$.

Conway's explanation towards a solution is as follows:

However, I have the following questions regarding the explanation above:

• Why $\{z:\Im(z,-i,0,i)>0\}=\{z:\Im(iz)>0\} ?$
• How does $T:=R^{-1}\circ S$ map the imaginary axis onto the circle $\Gamma\subset\mathbb C_\infty ?$
• How did he arrive at the expression $g(z)=\dfrac{e^z -1}{e^z+1}$ from the expression for $Tz$ ?

I understood his calculations other than these three conclusions.

Answer 1

I don't quite understand Conway's notation, but, this should be pretty straightforward using Cayley's transform and a rotation.

That is, $$f(z)=\frac{z-i}{z+i}$$ maps the upper half plane onto the unit circle.

So you just need to compose with a rotation by $\pi/2$. That's multiplication by $i$.

The result is $$\frac {z-1}{z+1}$$.