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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?

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$T(z)=\frac {(1+z)(1-\overline {z})} {|1-z|^{2}}$ 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 $\{z: |z|<1\}$.

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