= Solution
{c}
For $z=x+iy$,
$$
\operatorname{Im}g(z)=y\left(1-\frac1{x^2+y^2}\right).
$$
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 $\operatorname{Im}\zeta>0$. The equation $g(z)=\zeta$ is
$$
z^2-\zeta z+1=0.
$$
Its 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 $\operatorname{Im}g$ forces $y>0$, so it belongs to the fluid region. Moreover, if $g(z_1)=g(z_2)$, then
$$
(z_1-z_2)\left(1-\frac1{z_1z_2}\right)=0.
$$
Two distinct roots in the fluid region would require $z_1z_2=1$, 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>.
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