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I am now dealing with a math problem. It requires to prove the complex function $f(z)=\frac{1+z}{1-z}$ transforms the region inside a unit circle on the $z$ plane, $|z|<1$, to the right half of the $f(z)$ plane, $\Re f(z)>0$.

I tried to turn this into a solution solving question, which is to show that for any point $(a,b), a>0$ in the $f(z)$ plane, there is a analytic solution so that one can express $(a,b$) by $\Re z$ and $\Im z$ respectively. However, I can not solve the equations for $a$ and $b$.

I also tried to plug in $z=x+iy$ and analyze the result form, but failed to show it was the entire right half plane.

So I do not know how to deal with this.

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Note that if $\lvert z\rvert<1$, then\begin{align}\operatorname{Re}\left(\frac{1+z}{1-z}\right)&=\operatorname{Re}\left(\frac{(1+z)\left(1-\overline z\right)}{(1-z)\left(1-\overline z\right)}\right)\\&=\operatorname{Re}\left(\frac{1+z-\overline z-\lvert z\rvert^2}{\lvert 1-z\rvert^2}\right)\\&=\frac{1-\lvert z\rvert^2}{\lvert 1-z\rvert^2}\\&>0.\end{align}Therefore, the range of $f$ is contained in the right half-lane of $\mathbb C$.

In order to prove that the range of $f$ is actually equal to that half-plane, take $w$ from the half-plane, solve the equation $\frac{1+z}{1-z}=w$ and prove that the solution has absolute value smaller than $1$.

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  • $\begingroup$ Thank you! I did not think about that. $\endgroup$ – OneMoreSec Jul 17 '19 at 10:14
  • $\begingroup$ I'm glad I could help. $\endgroup$ – José Carlos Santos Jul 17 '19 at 10:15
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Hints: $\Re (\frac {1+z}{1-z})=\Re \frac {(1+z)(1-\overline {z})} {|1-z|^{2}}$ and this number is $>0$ when $|z| <1$. So $f$ maps the unit disk into the right half plane. Given $\zeta =a+ib$ with $a>0$ let $z=\frac {\zeta - 1}{\zeta +1}$ and verify that $|\zeta | <1$. [This is just the inequality $|a+ib -1|^{2} <|a+ib+1|^{2}$ or $(a-1)^{2}+b^{2} <(a+1)^{2}+b^{2}$]. Then show that $f(\zeta)=z$.

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