Finding the images under a Mobius transformation

Question

Let $L1$ be the $x$-axis, let $L2$ be the $y$-axis and let $L3$ be the vertical line $x = 1$. For each $k ∈ Z$ let $C_{k}$ denote the circle of radius $r = 1/2$ with centre $z = 1/2 + ki$. Let $f(z) = {2z}/{(z+1)}$.

1) Sketch $L1, L2, L3$ and the circles $C_{k}$.

2) Show that $f(z)$ maps $L1$ to itself and find the images $f(L2)$, $f(L3)$ and $f(C_{0})$ under the transformation $f(z)$.

I have attempted to sketch the points, I believe I should end up with circles one above each other but not sure?

I'm struggling with part 2) too. I'm told as a hint to consider the angles between $L1, L2, L3$ and $C_{0}$ but don't know how to use this. In my attempt I have just subbed say $z=iy$ for $L2$ to represent the y axis. I then did the same for 3 points on the circle to end up with $f(1/2)=2/3, f(-1/2) = -2$ and $f(i/2) = 2i/5+1/5$ but I have no idea whether this is right or where I would go from here as finally I am asked to produce another sketch of all of the images of the points without any further calculations.

Answer 1

For the circle $C_0$ I think that the approach you have taken is correct, by looking at $3$ points on the circle and studying where they get mapped.
But note that the points you chose are not on $C_0$.

$C_0$ is the circle around $z=1/2$ with radius $r=1/2$.
Let's look at $$z_1 = 1,\ \ z_2=\frac{1}{2} + \frac{i}{2}, \ \ z_3 = 0$$ under $f(z) = \frac{2z}{z+1}$ we have: $$ f(z_1) = 1, \ \ f(z_2) = \frac{4}{5}-\frac{2}{5}i, \ \ f(z_3) = 0$$

Since we know that Mobius transformations map circles and lines to circles or lines, we can check that these $3$ points lie on the same $C_0$ circle around $z=1/2$ with radius $r=1/2$.

I hope this helps.

Update:
To show that $f(z_1),\ f(z_2),\ f(z_3)$ lie on $C_0$ we can follow the steps in https://math.stackexchange.com/a/213670/201454

We get the following $3$ equations: $$(1-x_c)^2 + y_c^2 -r^2 = 0$$ $$(\frac{4}{5}-x_c)^2 + (-\frac{2}{5} - y_c)^2 -r^2 = 0$$ $$x_c^2 + y_c^2 -r^2 = 0$$

Subtracting the third equation from the first gives: $$1 - 2x_c = 0 \implies x_c=\frac{1}{2}$$

Subtracting the third equation from the second gives: $$\frac{16}{25}-\frac{8}{5}x_c + \frac{4}{25} + \frac{4}{5}y_c = 0 \implies \frac{4}{5} - \frac{8}{5}x_c + \frac{4}{5}y_c = 0$$

Plugging in $x_c=1/2$ gives $y_c=0$

Now plug in $x_c=1/2,\ y_c=0$ in the third equation to get $r=1/2$

Answer 2

(partial answer)

Let us consider the image of $(L_2)$ by function $f$.

The idea is to describe the "generic element" $z \in (L_2)$ as being:

$$z=i t , \ \ \ t \in \mathbb{C}, \ \ \text{giving} \ \ \ f(it)=\frac{2it}{it+1}$$

Let us now convert this expression in order that the denominator doesn't contain any more a complex non-real number. This is done by multiplying the numerator and the denominator of this fraction by the conjugate expression of the denominator, i.e.,

$$f(it)=\frac{2it(-it+1)}{(it+1)(-it+1)}=\frac{2t^2+i2t}{1+t^2}=x+iy \ \ \text{with} \ \ \cases{x=\frac{2t^2}{1+t^2}\\y=\frac{2t}{1+t^2}}$$

or in a more interesting manner, by a $-1$ translation on the $x$ axis:

$$\cases{x-1=\frac{1-t^2}{1+t^2}\\y=\frac{2t}{1+t^2}}$$

These are the well known parametric equations for the unit circle (see this). All the unit circle is reached but one point. Could you find it?

Thus the looked for image of $(L_2)$ is circle is with center $(1,0)$ and radius $1$.

Indeed one checks that $(x-1)^2+y^2=1$;

But in fact there is a missing point which is $(0,2)$ which is never "reached".

Edit: Circle $(C_0)$ is parametrically described as being the set of complex numbers of the form: $$z=\frac12+\frac12 e^{it}=\frac12+\frac12(\cos(t)+i \sin(t)).$$

Thus, the image of circle $(C_0)$ by transformation $f$ is thus the set of points of the form:

$$f(z)=\dfrac{(1+\cos(t))+i \sin(t)}{(\frac32+\frac12 \cos(t))+i (\frac12 \sin(t))}$$

You just have to multiply, as before, the numerator and the denominator by the conjugate of the denominator.