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I am trying to find a conformal mapping from the unit disk minus a line segment, e.g. $\mathbb{D}-\{z=x+i0:-1/2\leq x\leq1/2\}$, to an annulus. As far as I know the function $z\to z+1/z$ maps the annulus $\{z:1<|z|<2\}$ to an ellipse minus a line segment. But how can I get one that the outer boundary is a disk?

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Let $D=\mathbb{D}-\{z=x+i0:-1/2\leq x\leq1/2\}$, $D^+=\{z\mid z\in D,\, \operatorname{Im } z>0\}$ and $D^-=\{z\mid z\in D,\, \operatorname{Im }z <0\}$. We know that $$\zeta=\varphi (z)=\left(\frac{1+z}{1-z}\right)^2$$ maps $D^+$ conformally onto the upper half plane $H^+$. Note that $\varphi (-1/2)=1/9,$ $ \varphi (1/2)=9.$

By the Schwarz-Christoffel formula the function $$ \xi=\phi(\zeta) =A\int_0^\zeta \frac{dt}{\sqrt{t(9t-1)(t-9)}} $$ maps $H^+$ onto a rectangle with vertices $\xi=0,\, \xi=a>0,\, \xi=a+\pi i$ and $ \xi=\pi i$ for a suitable constant $A$. Therefore $$ w=f(z)=\exp(\phi(\varphi (z))$$ maps $D^+$ conformally onto the upper half $E^+$ of an annulus $E=\{w\mid 1<|w|<e^a\}$. Let $I=(-1,-1/2),\, J=(1/2,1).$ The image of $I$ by $f$ is $(1,e^a)$ and the image of $J$ is $(-e^a, -1)$.

Define $$ F(z)=\left\{ \begin{array}{ll} f(z) , & \text{if } z\in D^+\cup I\cup J \\ \overline{f(\overline{z})} ,& \text{if } z\in D^- . \end{array}\right. $$ Then by the Schwarz reflection principle $F(z)$ maps $D$ conformally onto $E$ . See Fig1. Fig1

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