Möbius transformation of an area between two lines

Question

$\DeclareMathOperator{\Re}{Re}$I wish to find where all the points in $$\left\{ z:\,\,\,-1<\Re\left(z\right)<0\right\} $$ are mapped to under $$f\left(z\right)=\frac{1}{z+1}$$ I found out $\Re\left(z\right)=0$ goes to a circle with a radius $1/2$ centered at $1/2$ and that $\Re\left(z\right)=-1$ goes to $\Re\left(z\right)=0$ but I have no clue on how proceed, how do I know where all the points inside are mapped to?

Answer 1

$f$ is a Möbius transformation, in particular a continuous automorphism of the extended complex plane $\hat {\Bbb C}$.

$f$ maps the extended line $L_1 = \{ z : \operatorname{Re}(z) = 0 \}\cup \{ \infty \} $ onto the circle $C = \{ z : |z-\frac 12| = \frac 12 \}$, and the extended line $L_2 = \{ z : \operatorname{Re}(z) = -1 \}\cup \{ \infty \} $ onto $L_1$.

Continuous functions map connected sets to connected sets, therefore each connected component of $\hat {\Bbb C} \setminus (L_1 \cup L_2)$ is mapped onto a connected component of $\hat {\Bbb C} \setminus (C \cup L_1)$.

Since $f(-1/2) = 2$, the region between $L_1$ and $L_2$ is mapped onto the intersection of the right half-plane with the exterior of the circle $C$.