 Solution

For the Joukowski map $F(z)=z+b^2/z$, the derivative is $F^{\prime }(z)=1-b^2/z^2$. Thus it is a conformal map at every finite $z$ except its pole at zero and its critical pointsThe pole maps to $\infty$; as a map of Riemann spheres it is a simple pole and is not a branch point. Neither is the point at infinity, where the map behaves like $z$.

An appropriate domain is the upper exterior half-discFor $z=x+iy$, $\mathrm{Im}F(z)=y(1-b^2/|z{|}^2)>0$ on $D$. To prove surjectivity and injectivity, solve $F(z)=\zeta$ as $z^2-\zeta z+b^2=0$. Its two roots have product $b^2$. For $\mathrm{Im}\zeta >0$, neither root lies on the real axis or on $|z|=b$, since either boundary maps to the real axis. Exactly one root therefore has modulus greater than $b$. The imaginary-part identity forces that root to have positive imaginary part. It is the unique root in $D$, so $F$ maps $D$ conformally and bijectively onto the upper half-plane.