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