# Show that there is a linear rational transformation

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?

• Hint: If $g(b) = \infty$ for a linear rational transformation $g$, what do you know about $g(L)?$ – Brian Moehring Aug 28 '19 at 3:39
• – Martin R Aug 28 '19 at 6:50

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.
• 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? – J.rafa Aug 30 '19 at 19:21
• @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. – Christian Blatter Aug 30 '19 at 19:32