= Solution
For the <Joukowski map> $F(z)=z+b^2/z$, the <derivative> is $F'(z)=1-b^2/z^2$. Thus it is a <conformal map> at every finite $z$ except its pole at zero and its critical points
$$
\boxed{z=\pm b,\qquad \zeta=F(z)=\pm2b.}
$$
The 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-disc
$$
\boxed{D=\{z:\operatorname{Im}z>0,\ |z|>b\}.}
$$
For $z=x+iy$, $\operatorname{Im}F(z)=y(1-b^2/|z|^2)>0$ 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 $\operatorname{Im}\zeta>0$, 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 \b[$F$ maps $D$ conformally and bijectively onto the upper half-plane].
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