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Let $$f(z)=\frac{z-a}{z-b}$$ with $a,b\in D(0,r)$ and $r>0$. Let $$E=\{z\in\mathbb C: f(z)\notin N\}$$

$$N=\{Re(z)\leq 0;Im(z)=0\}$$

How can i find $E$ in terms of $r$?

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Excluding the trivial case $a=b$ (when $E=\mathbb C$), the function $f$ is a bijection on the Riemann sphere $\widehat {\mathbb C}$. Therefore, $f^{-1}(N^c)=f^{-1}(N)^c$ where $f^{-1}(w)=\dfrac{bw-a}{w-1}$. Since fractional linear transformations map "circles" to "circles", the image of $N$ under $f^{-1}$ is the "circular" arc with endpoints $f^{-1}(0)=a$ and $f^{-1}(\infty)=b$, also passing through the point $f^{-1}(-1)=(a+b)/2$. Which is actually the line segment with endpoints $a$ and $b$.

(I'm using "circle" to mean something that is either a circle or a line.)

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