Let $C$ be a circle in the complex plane $\mathbb{C}$ whose centre is $(0,ir)$ (where $r$ is a non-zero positive real number) and is tangent to the real axis at the origin $(0,0)$ (the radius of the circle $C$ is obviously $r$).
Let $f: \mathbb{C} \rightarrow \mathbb{C}$ be a function defined by $$\omega \mapsto f(\omega):= \frac{\omega}{1-\omega}$$
Let $p=(x,y) \in C$ (different from the origin $(0,0)$).
Is $f(p) \in C$? (if yes, how to show that?).
Thanks in Advance.
On
The equation of the circle $C$ is $$|z-ir|=r$$
Let $\omega$ be the image of the point $z\in C$ under the transformation given by $f$.
Then $$w=\frac{z}{1-z}\implies z=\frac{\omega}{1+\omega}$$
So $\omega$ satisfies the equation $$\left|\frac{\omega}{1+\omega}-ir\right|=r$$ $$\implies|w-ir(1+\omega)|=r|1+\omega|$$
Now write $\omega=u+iv$, and this gives $$|(u+rv)+i(v-r-ru)|=r|1+u+iv|$$
$$\implies(u+rv)^2+(v-r-ru)^2=r^2\left((1+u)^2+v^2\right)$$
Simplifying this leads to $$u^2+v^2-2vr=0$$ and this is indeed the equation of $C$, so $f$ maps $C$ onto itself.
Here is a geometrical approach.
One can write the given map $f$ as a composition: $$z\to-z\to(1-z)\to\frac 1{1-z}\to \frac 1{1-z}-1\ .$$ Let us see what happens with the given circle after each map.
$z\to -z=z_1$, reflection w.r.t. origin, it maps the given circle $\mathcal C=C(ir,r)$ (of center $ir$ and radius $r$) into the circle $\mathcal C_1=C(-ir, r)$ tangent in $0$ to the $Ox$-axis.
$z_1\to 1+z_1=z_2$, a translation by one, maps $\mathcal C_1=C(-ir,r)$ to $\mathcal C_2=C(1-ir,r)$, a circle tangent in $1$ to the $Ox$-axis.
$z_2\to\frac 1{z_2}=z_3$ is an inversion $*$ of power one, followed by a reflection. The inversion brings $z_2=\rho(\cos t+i\sin t)$ to $\frac 1\rho(\cos t+i\sin t)$, and after it, we reflect w.r.t. the $Ox$ axis to get the final result $\frac 1\rho(\cos t-i\sin t)=\frac 1{\rho(\cos t+i\sin t)}$.
What happens with the circle $\mathcal C_2$ when we apply the inversion only? The point $A=1$ of this circle is kept in place. The circle $\mathcal C_2$ is mapped also into a circle $\mathcal C_2^*$, tangent to $A^*=1^*=1=A$ to the $Ox^*$-ray, which is the $Ox$-ray. We know one more point of $\mathcal C_2$, which is $B=1-2ir$.
This point goes to a point $C=B^*=(1-2r)^*=t(1-2ir)$ on the line from $0$ to $(1-2ir)$, $t\in \Bbb R_{>0}$, such that the product of the absolute values is one. So $t=\frac 1{|1-2ir|^2}=\frac 1{1+4r^2}$, if we really want to compute it. But we don't. Let us forget this, was added only for the sake of figuring out what the inversion is doing with affixes.
But geometrically we have a better understanding. Where is $C=B^*$? By the properties of the inversion, $OA^2=OB\cdot OC=1$, and $\Delta OAB\sim\Delta OCA$. So $AC\perp OB$, the triangle $\Delta ABC$ has a right angle in $C$, and $C$ is on the same circle $\mathcal C_2$, with center in the mid point $1-ir$ of $AB$. This means that $\mathcal C_2^*=\mathcal C_2$, the inversion invariates it.
We still need to reflect $\mathcal C_2$ w.r.t. the $Ox$ axis, so we obtain the circle $\mathcal C_3=\bar {\mathcal C}_2=\overline{C(1-ir, r)}=C(\overline{1-ir}, r)=C(1+ir,r)$.
Finally, the map $z_3\to z_3-1=z_4$ (translation) brings $\mathcal C_3=C(1+ir, r)$ into $C(1+ir-1,r)=C(ir,r)=\mathcal C$, the initial circle.
$\square$
Alternative proof (Möbius and an one point check):
The given function is a Möbius transformation, a conformal map. It is known that the class of all lines and circles is mapped to itself. So the image of $\mathcal C=C(ir, r)$ is either a circle or a line. The circle is tangent in $0$ to the $Ox$ axis. The image $f(\mathcal C)$ is then tangent in $f(0)=0$ to the transform $f(Ox)=Ox$. (Real points go to real points.) So it is a circle through $0$ (cannot be $Ox$), its center is on $Oy$, and we need one more point on it. For example $f(2ir)=\frac {2ir}{1-2ir}=\frac 1{1+4r^2}2ir(1+2ir)$. Is this point also on $\mathcal C$? Let us compute its distance to $ir$: $$ \begin{aligned} d(ir,f(2ir)) &=\left|ir-\frac 1{1+4r^2}2ir(1+2ir)\right| =\frac 1{1+4r^2}|ir|\cdot|(1+4r^2)-2(1+2ir)| \\ &=\frac r{1+4r^2}\cdot|(-1+4r^2)-4ir| =\frac r{1+4r^2}\cdot\sqrt{(1-8r^2+16r^2)+16r^2} \\ &=r \ . \end{aligned} $$