Let \(\DD\cap\HH=\ts{z\st\abs{z}<1,\ \Im z>0}\) be the upper half-disc and \(\HH=\ts{w\st\Im w>0}\) the upper half-plane. The map \[ F\colon\DD\cap\HH\to\HH,\qquad F(z)=-\frac12\big(z+z^{-1}\big), \] is a biholomorphism.
For \(z=re^{i\theta}\) with \(0<r<1\) and \(0<\theta<\pi\), \[ F(z)=-\frac12\Big(r+\frac1r\Big)\cos\theta+\frac i2\Big(\frac1r-r\Big)\sin\theta, \] so \(\Im F(z)>0\) and \(F\) is a holomorphic map into \(\HH\). Given \(w\in\HH\), the equation \(F(z)=w\) is \(z^2+2wz+1=0\). Its two roots have product \(1\), and neither lies on the unit circle or the real axis, because \(F\) maps the unit circle into \([-1,1]\) and \(\RR\sm\ts{0}\) into \(\RR\). So exactly one root \(z\) satisfies \(\abs{z}<1\), and it satisfies \(\Im z>0\) because \(\Im F(z)\) has the sign of \(\sin\theta\) when \(\abs{z}<1\). Hence \(F\) is bijective, and a bijective holomorphic map has a holomorphic inverse.
The map \(z\mapsto\frac12(z+z^{-1})\) is the Joukowski map; \(F\) is its negative. On the boundary of the half-disc, \(F\) maps the upper unit semicircle onto \([-1,1]\), with \(F(1)=-1\), \(F(i)=0\), \(F(-1)=1\), and it maps the segments \((0,1)\) and \((-1,0)\) onto \((-\infty,-1)\) and \((1,\infty)\) respectively. The inverse is \(F^{-1}(w)=-w+\sqrt{w^2-1}\), with the branch of the square root for which this value lies in \(\DD\).
The Joukowski map \(J(z)=\frac12(z+z^{-1})\) satisfies \(J(z)=J(1/z)\) and maps each of the punctured disc \(\ts{z\st 0<\abs{z}<1}\) and the exterior \(\ts{z\st\abs{z}>1}\) conformally onto \(\CC\sm[-1,1]\). It maps the upper half-disc onto the lower half-plane, and each circle \(\abs{z}=r\) with \(r\neq1\) onto the ellipse with foci \(\pm1\) and semi-axes \(\frac12(r+r^{-1})\) and \(\frac12\abs{r-r^{-1}}\) [@Ahl79].