Let $f: \mathbb{C} \to \mathbb{C} $ be given by $$f(z) = \frac{2iz}{z + i}$$,
$C$ be an arc of a circle passing through $0$ and $i$, and $$ A = \left\{ z \in \mathbb{C} : \text{Re}(z) > 0 \land \left| z - \frac{i}{2} \right| < \frac{1}{2} \right\} $$
Determine whether the following proposition is false or true and provide a proof for your assertion: $f(C) \subseteq C$ and $f(A) = A$.
My first aproach was to consider $z = r \cdot e^{i\theta} + a + bi$ as a random circle on the complex plane with center at $a+bi$, and since $C$ pass for $0$ and $i$ the distance from the center at any point of $C$ is $\frac{1}{2}$ representing a parametric $f(z)$
$$z = \frac{1}{2} e^{i\theta} + \frac{i}{2}$$ $$f(z(\theta)) = \frac{2i \left( \frac{1}{2} e^{i\theta} + \frac{i}{2} \right)}{\frac{1}{2} e^{i\theta} + \frac{i}{2} + i}$$
but idont know where to go from here