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