 Solution

For $z=x+iy$,Both factors are positive in the fluid region, so its image is contained in the upper half-plane. The boundary correspondence is also explicit: on the upper unit semicircle, $z=e^{i\theta }$ gives $g(z)=2\cos \theta \in [-2,2]$, while the real rays $x>1$ and $x<-1$ map onto the rays $\zeta >2$ and $\zeta <-2$.

To prove the image is the whole upper half-plane, fix any $\mathrm{Im}\zeta >0$. The equation $g(z)=\zeta$ isIts roots have product one. Neither root can have unit modulus, since that would make $\zeta =z+1/z$ real. Exactly one root therefore has modulus greater than one. For that root the formula for $\mathrm{Im}g$ forces $y>0$, so it belongs to the fluid region. Moreover, if $g(z_1)=g(z_2)$, thenTwo distinct roots in the fluid region would require $z_1{z}_2=1$, impossible when both moduli exceed one. Thus the Joukowski map of the upper exterior of a unit disk is a bijection from the fluid region onto the upper half-plane.