\[{H^{(1)}_{2/3}}\left(\zeta\right)=e^{-2\pi i/3}{H^{(1)}_{-2/3}}\left(\zeta%
\right)=e^{\pi i/6}(\sqrt{3}/z)\left(\operatorname{Ai}'\left(-z\right)-i%
\operatorname{Bi}'\left(-z\right)\right),\]