= Square-root Loewner driving function generates a straight slit
For the <Chordal Loewner equation> driven by $U_t=a\sqrt t$, the hull is a straight slit whose endpoint is proportional to $\sqrt t$. Put $r_\pm=(a\pm\sqrt{a^2+16})/2$ and $\alpha=-r_-/(r_+-r_-)\in(0,1)$. Its inverse map is
$$
f_t(w)=(w-r_+\sqrt t)^\alpha(w-r_-\sqrt t)^{1-\alpha},
$$
using upper-half-plane branches. The interval between the two real roots maps onto the two sides of a segment of argument $\pi\alpha$.
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