# How to determine the range of a complex function

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.

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 $$1$$.
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 $$f(\zeta)=z$$.