= Solution
The two approaches to the slit base are distinct <boundary> points, each with interior <angle> $\pi/2$, whereas the slit tip has interior <angle> $2\pi$. Their exponents in the <Schwarz-Christoffel formula> are therefore $-1/2$, $1$ and $-1/2$. Write $a=x_2\in(-1,1)$ and take
$$
S(z)=\sqrt{z^2-1}
$$
on the <upper half-plane>, with the branch normalized by $S(z)\sim z$ at <infinity>. The <derivative> has the form
$$
f'(z)=c\frac{z-a}{S(z)}.
$$
On $-1<x<1$, the upper <boundary value> is $S_+(x)=i\sqrt{1-x^2}$. Both endpoints of this interval map to the slit base, so
$$
0=f(1)-f(-1)=c\int_{-1}^1\frac{x-a}{i\sqrt{1-x^2}}\,dx=i\pi ca.
$$
The odd term integrates to zero, and $c\ne0$; hence $\boxed{x_2=0}$. Now $S'(z)=z/S(z)$, so $f=cS+b$. The endpoint values force $b=0$, and the tip value, using $S(0)=i$ as a <boundary> limit, forces $c=1$. Thus
$$
\boxed{f(z)=\sqrt{z^2-1},\qquad f(z)\sim z\text{ at infinity}.}
$$
The branch is important: on $x>1$ its <boundary value> is positive real, on $x<-1$ negative real, and on $(-1,1)$ positive imaginary. The intervals $(-1,0)$ and $(0,1)$ traverse the two sides of the slit, meeting at its tip.
The map is injective in the <upper half-plane>: equal images imply $z_1^2=z_2^2$, and the alternative $z_1=-z_2$ cannot put both points in that half-plane. Its <derivative> has no interior zero. The image has positive <imaginary part>, by <continuity> of the branch and the fact that an interior real image would require $z^2-1$ to be nonnegative real, which is impossible for $\operatorname{Im}z>0$. Conversely, a point $\zeta$ in the <upper half-plane> outside the slit gives a unique upper-half-plane root of $z^2=\zeta^2+1$. A point $\zeta=iy$ with $0<y\leq1$ gives only real roots or zero and is excluded. These observations verify that the image is exactly the required slit domain, rather than just matching its <boundary>.
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