# which domain does this Möbius map to $\Re(w) > 0$?

Given the following Möbius: $$w = T(z) = \frac{1+z}{1-z}$$ How could I find the domain of $$Z$$ which $$T$$ maps to $$\{\Re(w)>0\}$$?

I tried to inverse $$T$$ and got: $$z = T^{-1}=\frac{w-1}{w+1}$$

Then I tried to see where the reversed Möbius maps the imaginary axis and got: $$T^{-1}(i)=-\sqrt{2}i$$ $$T^{-1}(0)=-1$$ $$T^{-1}(-i)=\sqrt{2}$$

But then I could not conclude anything! How could I find out the mapped domain?

$$T(z)=\frac {(1+z)(1-\overline {z})} {|1-z|^{2}}$$ so the real part of $$T(z)$$ is $$\frac {1-|z|^{2}} {|1-z|^{2}}$$. This is positive iff $$|z| <1$$ so the answer is $$\{z: |z|<1\}$$.