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 $\{z: |z|<1\}$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.