Conformal mapping from the unit disk minus a line segment to an annulus

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?

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 $$E=\{w\mid 1<|w|. Let $$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