\[\operatorname{Bi}\left(-z\right)=\sqrt{z/3}\left(J_{-1/3}\left(\zeta\right)-J_%
{1/3}\left(\zeta\right)\right)=\tfrac{1}{2}\sqrt{z/3}\left(e^{2\pi i/3}{H^{(1)%
}_{1/3}}\left(\zeta\right)+e^{-2\pi i/3}{H^{(2)}_{1/3}}\left(\zeta\right)%
\right)=\tfrac{1}{2}\sqrt{z/3}\left(e^{\pi i/3}{H^{(1)}_{-1/3}}\left(\zeta%
\right)+e^{-\pi i/3}{H^{(2)}_{-1/3}}\left(\zeta\right)\right),\]