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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?

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    The map $f$ is conformal, and (as we discussed in another of your questions:) a composition of translations and inversion in the unit circle, which therefore maps (circles or lines) to (circles or lines). I haven't tried to sketch your loci in detail, but the preceding geometric observations are presumably enough to specify a unique (circle or line) for each circle. – 2017-01-14
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    @AndrewD.Hwang I'm confused as we have not been taught what conformal means. A google search returns 'angle-preserving' but if that is the case, as $f(z)$ maps $L_1$ to itself, it must surely fix $L_2,L_3$ and $C_0$? – 2017-01-15
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    Conformality only tells us that $f(L_{2})$ meets $f(L_{1}) = L_{1}$ at a right angle; lots of curves do this. :) Since $f(z) = 2 - \frac{2}{z + 1}$, the action of $f$ can be broken into four steps: (i) Translate by $1$ ($z \mapsto z + 1$); (ii) Invert in the unit circle and conjugate ($z \mapsto 1/z$); (iii) Scale by $-2$ ($z \mapsto -2z$); (iv) Translate by $2$. For each of your curves, perform these steps in order (sketching the end result on a single diagram) and see if the result makes intuitive sense. – 2017-01-15
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    @AndrewD.Hwang I've done what I can. I get $f(L_2)$ is the unit circle, $r=1$, centre $=(1,0)$. I don't know how to find $f(L_3)$. $f(C_0) = C_0$. Does this sound right? – 2017-01-16
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    There was no bounty when I started composing my answer, and I hope my answer doesn't detract others from writing something more helpful or detailed. – 2017-01-16

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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, $f(C_{0}) = C_{0}$.

Images of circles under a rational function