= Solution
On $\zeta=ae^{i\theta}$, the <Joukowski map> gives $z=2a\cos\theta$, so the image is the segment $S=[-2a,2a]$. Solving the resulting quadratic gives the inverse candidates
$$
\zeta=\frac12\left(z\pm\sqrt{z^2-4a^2}\right).
$$
On $D=\mathbb C\setminus[-2a,2a]$, choose the analytic square <root> $s(z)$ of $z^2-4a^2$ characterized by $s(z)/z\to1$ at infinity. One explicit definition is
$$
s(z)=(z+2a)\sqrt{\frac{z-2a}{z+2a}},
$$
where the square <root> on the right is the <principal square root>. Its argument never lies on the nonpositive real axis for $z\in D$, so this defines the required <holomorphic function>. Then $g(z)=(z+s(z))/2\sim z$.
The two quadratic roots have product $a^2$. For $z\notin S$, neither <root> lies on $|\zeta|=a$, since that circle maps into $S$. Consequently exactly one <root> lies outside the circle, and it is the branch $g$ chosen at infinity. Also, if two exterior points have the same image, then
$$
(\zeta_1-\zeta_2)\left(1-\frac{a^2}{\zeta_1\zeta_2}\right)=0,
$$
whose second factor cannot vanish. Thus the <Joukowski map> is a <conformal map> from $|\zeta|>a$ onto $D$, and $g$ is <conformal map>[conformal] throughout $D$. The two endpoints are branch points and are excluded.
The <complex potential of a two-dimensional potential flow>[complex potential] in the circle plane is the sum of the <uniform flow> term $Ue^{-i\alpha}\zeta$, its circle-image term $Ua^2e^{i\alpha}/\zeta$, and the <point vortex> term $-i\Gamma\log\zeta/(2\pi)$. On $\zeta=ae^{i\theta}$, the first two sum to $2Ua\cos(\theta-\alpha)$, which is real, while the imaginary part of the vortex term is the constant $-\Gamma\log a/(2\pi)$. Hence the circle is a streamline, giving no normal flow. At infinity the first term gives the required <uniform flow>; the logarithmic term supplies circulation $\Gamma$. The potential may be multivalued by a real constant, but its velocity is single-valued.
Conformal transport therefore gives
$$
\boxed{w(z)=U\left(g(z)e^{-i\alpha}+\frac{a^2e^{i\alpha}}{g(z)}\right)
-\frac{i\Gamma}{2\pi}\log g(z).}
$$
In the frame where the plate is horizontal, its length is $4a$ and the stream has angle $\alpha$; rotating the frame gives the equivalent incidence description. The <complex velocity> is
$$
\frac{dw}{dz}=\frac{W'(\zeta)}{1-a^2/\zeta^2}.
$$
At the trailing edge $z=2a$, corresponding to $\zeta=a$, the denominator has a simple zero. Finite velocity requires
$$
0=W'(a)=-2iU\sin\alpha-\frac{i\Gamma}{2\pi a},
$$
so the <Kutta condition> is
$$
\boxed{\Gamma=-4\pi Ua\sin\alpha.}
$$
With this choice, cancellation of the common zero gives
$$
\left.\frac{dw}{dz}\right|_{z=2a}
=\frac{W''(a)}{2/a}
=Ue^{i\alpha}+\frac{i\Gamma}{4\pi a}
=\boxed{U\cos\alpha}.
$$
Thus the limiting velocity has zero transverse component and is tangent to the plate.
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