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Let $L$ be a circle in $\mathbb{C}$ and $a,b\in\mathbb{C}\setminus{L}$ . It shows that there is a linear rational transformation $f$ such that $L\cup{\{a}\}$ is contained in the domain of $f$, $f (a) = b$ and $f (L) = L$.

Can anybody help me?

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In the framework of complex linear transformations (Moebius transformations) we may assume that $L$ is the real axis, and that $a$, $b\notin{\mathbb R}$. Since we want $f(\bar{\mathbb R})=\bar{\mathbb R}$ and $f(a)=b$ we also need $f(\bar a)=\bar b$. Going through the computation we find out that there are many such $f$s. Therefore we impose the additional constraint $f(\infty)=\infty$. In other words, we are looking for an $$f(z)=\lambda z+\mu$$ satisfying the given conditions. The coefficients $\lambda$ and $\mu$ are found by solving the system $$b=\lambda a+\mu,\qquad\bar b=\lambda\bar a+\mu\ .$$ This leads to $$\lambda={b-\bar b\over a-\bar a}\in{\mathbb R},\qquad \mu={a\bar b-\bar a b\over a-\bar a}\in{\mathbb R}\ ,$$ so that $f(\bar{\mathbb R})=\bar{\mathbb R}$ is automatically fulfilled. This proves the existence of a Moebius transformation $f$ doing the desired job.

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  • $\begingroup$ Hi, thank you very much for answering, I have two question, why can we assume that $L$ is the real axis? and what does $\mathbb{\bar{R}}$ denote? $\endgroup$ – J.rafa Aug 30 '19 at 19:21
  • $\begingroup$ @Yessit: All circles (and lines) are alike in this world. But you could also argue as follows: There is a certain Moebius transformation $T$ mapping the given circle $L$ to the line ${\mathbb R}$. This $T$ moves $a$ and $b$ to new points $a'$, $b'\notin{\mathbb R}$. With $a'$ and $b'$ do what I have described; then move everything back. $\endgroup$ – Christian Blatter Aug 30 '19 at 19:32

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