Images of two arcs under $z+1/z$

Question

Let $C$ be the circle $|z-ai|=\sqrt{1+a^2}$, where $a>0$. Let $f(z)=z+\frac{1}{z}$, where $z\in \mathbb{C}$. Let $C_1, C_2$ be the two arcs on $C$ determined by $-1$ and $1$. I am looking for an easy way (or at least one way) to prove that $$f(C_1)=f(C_2)$$

Answer 1

Möbius transformations map circles and straight lines to circles and straight lines. $z \mapsto 1/z$ is a Möbius transformation, it fixes $1$ and $-1$, and maps $i(a - \sqrt{1+a^2})$ to $i(a + \sqrt{1+a^2})$, hence it maps $C_1$ to $C_2$ and vice versa. Since $f(1/z) = f(z)$, we have $f(C_1) = f(C_2)$ as desired.