 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 isusing upper-half-plane branches. The interval between the two real roots maps onto the two sides of a segment of argument $\pi \alpha$.