Complex geometry - image under transformation by considering angles I have asked this question elsewhere and was given a good approach to solving it by parametrising the curves. However, I have been told there is another method that can be used that involves comparing the angles between the curves.
Let $L_1$ be the $x$-axis, let $L_2$ be the $y$-axis and let $L_3$ be the vertical line $x=1$. For each $k \in \mathbb{Z}$ let $C_k$ denote the circle of radius $r=\frac{1}{2}$ with centre $z=\frac{1}{2}+ki$. Let $f(z)=\frac{2z}{z+1}$.
Find the images $f(L_2),f(L_3)$ and $f(C_0)$ by considering the angles between $L_1,L_2,L_3$ and $C_0$.
I have sketched this out and I get a circle centred at $(\frac{1}{2},0)$ with $r=\frac{1}{2}$, with $L_2$ a tangent at $(0,0)$, $L_3$ a tangent at $(1,0)$ and $L_1$ being the diameter perpendicular to $L_2$. How would I use this information to determine the images?
 A: Hints: Your curves are three lines (one horizontal, two vertical) and a family of circles tangent to $L_2$ and $L_3$, with successive circles mutually-tangent. The image of the imaginary axis $L_2$ can be found in stages:


*

*Translate by $1$: The line $x = 1$;

*Invert in the unit circle and conjugate: The circle centered on the real axis (by symmetry) and crossing the axis at $1$ and $0$, i.e., the circle of radius $1/2$ centered at $(1/2, 0)$;

*Scale by $-2$: The circle of radius $1$ centered at $(-1, 0)$;

*Translate by $2$: The circle of radius $1$ centered at $(1, 0)$.
Generally, the image of the vertical line $x = c \neq 0$ under inversion in the unit circle is the circle whose center lies on the $x$-axis, and which passes through the points $(1/c, 0)$ and the origin. (Under inversion in a circle $C$, every line maps to a circle through the center of $C$.)
As for the circles $C_{k}$: Their images under $f$ are circles, each tangent to $f(L_2)$ and $f(L_3)$, and with $f(C_{k})$ tangent to $f(C_{k-1})$ and $f(C_{k+1})$. As you say, $f(C_{0}) = C_{0}$.

