What is the conformal equivalence from the half-plane to the unit disc? Problem: Find a conformal equivalence from the half-plane {z : Re(z) > 1} to the unit disc D. 
My Attempt: Just a FYI that I am completely new to conformal mapping. 
Okay, so I know that $$h(z) = \frac{1+z}{1-z}$$ is the conformal mapping from the unit disc to the right-half plane. If I take its inverse, do I get the conformal mapping from the RHP to the unit disc? In other words, is $$h^{-1}(z) = \frac{z-1}{z+1}$$ the conformal mapping from the RHP to the unit disc? If so, then how do I make it so that I get the conformal mapping from only the half plane {z : Re(z) > 1} to the unit disc? Do I simply translate it by adding 1? $$h^{-1}(z) = \frac{z-1}{z+1}+1$$
Please help! A step-by-step process would be absolutely amazing! Thank you!
 A: You may use this:
Let $H=\left\{z:Rez >1\right\} ,D=\left\{z: |z|<1\right\}.$ Take points $a=1-i,b=1,c=1+i ,A=-i,B=-1,C=i .$ Then $a,b,c\in \partial H ,A,B,C \in \partial D.$ Moving from $a$ to $c$,the right side of $\partial H$ is $H$ and moving form $A$ to $C$, the right side of $\partial D$ is $D.$ Therefore, by the orientation principle it is enough to find the Mobius transformation $T$ such that $T(a)=A,T(b)=B, T(c)=C.$
To do this,let $T_1$ be the mobius transformation such that $T_1(a)=1,T_1(b)=0,T_1(c)=\infty$ and $T_2$ be such that $T_2(A)=1,T_2(B)=0,T_2(C)=\infty$. But then (see Conway on Cross ratios)
$$T_1(z)=\frac{(z-b)(a-c)}{(z-c)(a-b)}$$
$$T_2(z)=\frac{(z-B)(A-C)}{(z-C)(A-B)}$$
Finally ,$T=T_2^{-1}\circ T_1$ will do what you want.
A: Yes, you'll get conformal mapping with $h^{-1}$. Translating by $1$ is a good idea, but in your case you will map the halfplane to the translated circle. Try using translation on $h$ and then computing the inverse.
Also you can just try to find the fractional linear transformation with pole on the line $Re(z)=1$ that maps this line to the unit circle (are you familiar with the properties of fractional-linear transformations?)
