= Solution
An <explicit disk map of a slit half disk> makes the two <boundary> approaches visible without assuming an extension theorem for the <boundary>. We construct the map as a succession of <bijections> and then determine its limits.
First $w=z^2$ maps the upper half-disc of radius two bijectively onto $|w|<4$ slit along $[0,4)$. Indeed, its inverse chooses the <square root> whose argument lies in $(0,\pi)$. The removed segment on the <imaginary axis> becomes $[-1,0)$, so the image of $D$ is precisely
$$
\Omega=\{w:|w|<4\}\setminus[-1,4).
$$
The <Möbius transformation> $v=4(w+1)/(w+16)$ maps the radius-four disc bijectively onto the <unit disc>, taking $-1$ to zero and $4$ to one. This follows also by writing it as $(w/4+1/4)/(1+w/16)$, an <automorphism of the unit disk> after the scaling. It takes the removed real interval to $[0,1)$, so the image is the unit disc with this radial slit removed. Choosing $0<\arg v<2\pi$ defines a single-valued <square root> $u=\sqrt v$ and maps that slit disc bijectively onto
$$
U=\{u:|u|<1,\ \operatorname{Im}u>0\}.
$$
All these maps have nonzero <derivatives> on their respective domains, since their potentially problematic zero or branch point is on the removed slit.
Next the <Joukowski map> $J(u)=u+1/u$ is a <conformal bijection> from $U$ onto the <lower half-plane>. For $u\in U$,
$$
\operatorname{Im}J(u)=\operatorname{Im}u\,(1-|u|^{-2})<0.
$$
If $J(u)=J(t)$, then $(u-t)(1-1/(ut))=0$, and $|ut|<1$ rules out the second factor vanishing. To prove surjectivity, take $q$ in the <lower half-plane> and consider $u^2-qu+1=0$. Neither root lies on the unit <circle>, where $u+1/u$ is real. Their product is one, so exactly one has modulus less than one. For that root the displayed imaginary-part identity forces $\operatorname{Im}u>0$. Thus it lies in $U$. Also $J'(u)=1-u^{-2}$ cannot vanish there. Finally $C(q)=(q+i)/(q-i)$ maps the <lower half-plane> bijectively onto the <unit disc>: the inequality $|q+i|<|q-i|$ is exactly $\operatorname{Im}q<0$, and its inverse is $q=i(\zeta+1)/(\zeta-1)$. Consequently
$$
F(z)=\frac{u(z)+u(z)^{-1}+i}{u(z)+u(z)^{-1}-i},\qquad
u(z)^2=\frac{4(z^2+1)}{z^2+16},\qquad 0<\arg u(z)<\pi,
$$
is a <conformal bijection> of $D$ onto the disc.
At the slit point $z_0=i/2$ the continuous rational expression for $v$ tends to $4/21$. In a sufficiently small neighborhood of $z_0$, points of $D$ have nonzero <real part>. Since $\operatorname{Im}(z^2)=2\operatorname{Re}z\operatorname{Im}z$ and the real-coefficient <Möbius map> $w\mapsto v$ preserves the sign of the <imaginary part>, right-side approach gives $\arg v\to0$, while left-side approach gives $\arg v\to2\pi$. Hence
$$
u(z)\longrightarrow\begin{cases}c=2/\sqrt{21},&\operatorname{Re}z>0,\\-c,&\operatorname{Re}z<0,\end{cases}
\qquad
J(u(z))\longrightarrow\begin{cases}L=25/(2\sqrt{21}),&\operatorname{Re}z>0,\\-L,&\operatorname{Re}z<0.\end{cases}
$$
The two limits of $F$ are therefore
$$
\gamma_+=\frac{L+i}{L-i},\qquad \gamma_-=\frac{-L+i}{-L-i}.
$$
They have modulus one and are distinct, since $L>0$ and the <Möbius map> $C$ is <injective> on the extended real line. The <two-sided cluster set at an interior slit point> is a geometric consequence of opening the slit into two <boundary> intervals.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2004/iii/paper-9-slit.png]
{title=Two sides of the slit at i/2 become distinct boundary points after squaring, a disk automorphism and a square root}
For the given map $f$, the composition $B=f\circ F^{-1}$ is an <automorphism of the unit disk>. To see its precise <boundary> behavior, let $a=B^{-1}(0)$ and compose $B$ with the inverse of $T_a(\zeta)=(\zeta-a)/(1-\overline a\zeta)$. This gives a <disc automorphism> fixing zero. Applying <Schwarz lemma> to it and its inverse shows it is a <rotation>. Thus
$$
B(\zeta)=e^{i\theta}\frac{\zeta-a}{1-\overline a\zeta},\qquad |a|<1.
$$
Its denominator never vanishes on the closed disc, and its inverse has the same property; it therefore extends to a continuous <bijection> of the unit <circle>. Put \b[$\alpha=B(\gamma_+)$ and $\beta=B(\gamma_-)$]. These are distinct points of that <circle>, independent of the approaching sequence.
For any sequence approaching $z_0$, every infinite right-side subsequence has image tending to $\alpha$, and every infinite left-side subsequence has image tending to $\beta$. At least one side occurs infinitely often. Conversely, any convergent image subsequence has a further subsequence confined to one side, so its limit must be the corresponding one of these two points. The cluster set is therefore \b[exactly $\{\alpha\}$, $\{\beta\}$ or $\{\alpha,\beta\}$], according to which sides occur infinitely often. Both sides really occur in the domain, for example along $i/2\pm\varepsilon_n$ with positive $\varepsilon_n\downarrow0$.
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