= Solution
{c}
Choose each factor with its argument between zero and $\pi$ in the <upper half-plane>. On an interval between consecutive real prevertices, each factor has a constant boundary argument, so $f'$ has constant argument and the image is a straight side. Crossing $X_j$ from left to right decreases $\arg(Z-X_j)$ by $\pi$, hence increases $\arg f'$ by $\beta_j\pi$, giving the prescribed exterior turn. Locally,
$$
f(Z)-z_j\sim C_j(Z-X_j)^{1-\beta_j},
$$
so an upper half-disc maps to angle $(1-\beta_j)\pi$, the required interior angle. The overall constant rotates and scales the polygon; integration adds translation. For a general prescribed polygon, the <prevertex> positions must also satisfy the side-length conditions: the <derivative> formula fixes angles but arbitrary choices do not fix arbitrary side lengths.
For the slit domain, the slit tip has interior angle $2\pi$, so its exterior parameter is $-1$. The two vertices at infinity give the factors at $\pm1$, leading to $f'(Z)$ proportional to $Z(1-Z^2)^{-3/2}$. Integrating and imposing the tip value gives
$$
\boxed{z=f(Z)=\frac1{\sqrt{1-Z^2}}.}
$$
The <square root> is the principal one: for $\Im Z>0$, $1-Z^2$ never lies on the nonpositive real axis. Thus $\Re f>0$. Boundary points $-1<X<1$ map to the two banks of the slit, with $X=0$ at the tip; the outside intervals map to the imaginary axis. The <derivative> is nonzero in the open <half-plane>. Conversely each point of the slit domain determines exactly one root $Z$ of $Z^2=1-z^{-2}$ with positive <imaginary part>, proving bijectivity.
The <Möbius map> $(Z-1)/(Z+1)$ takes the <upper half-plane> to itself, so its <principal argument> lies in $(0,\pi)$. The <imaginary part> of its analytic <logarithm> is <harmonic>, equals $\pi$ on $(-1,1)$ and zero on the other real intervals. Thus
$$
U(X,Y)=\frac1\pi\arg\frac{Z-1}{Z+1}.
$$
It lies between zero and one and has the required boundary limits away from the two discontinuities. Pulling this <harmonic function> back by the inverse <conformal map> yields
$$
\boxed{u(x,y)=\frac1\pi\arg\frac{Z(z)-1}{Z(z)+1},\qquad
Z(z)^2=1-z^{-2},\quad\Im Z(z)>0.}
$$
This specifies all branches and remains bounded even as $z$ tends to either end at infinity.
To exhibit nonuniqueness without the growth restriction, put
$$
H(z)=\Im\frac{Z(z)}{1-Z(z)^2}.
$$
This is <harmonic> in the domain. On its boundary the corresponding $Z$ is real and the rational expression is real, except at $\pm1$, which represent infinity. At $Z=\infty$, representing $z=0$, the expression tends to zero. Thus $H$ has zero finite boundary data. It is nonzero, since for $Z=iY$ its value is $Y/(1+Y^2)>0$, and it is unbounded as $Z$ approaches either pole $\pm1$ non-tangentially. Therefore \b[$u+cH$ gives distinct solutions] with the same finite boundary data when boundedness at infinity is removed.
\Image[/past-exam-of-the-perturbation-methods-and-applied-complex-variables-course-of-the-university-of-oxford/2007-slit-domain-map.png]
{title=Prevertices, the slit domain, and the bounded <harmonic> <boundary values>}
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