\[\operatorname{ber}_{\nu}x+i\operatorname{bei}_{\nu}x=J_{\nu}\left(xe^{3\pi i/4%
}\right)=e^{\nu\pi i}J_{\nu}\left(xe^{-\pi i/4}\right)=e^{\nu\pi i/2}I_{\nu}%
\left(xe^{\pi i/4}\right)=e^{3\nu\pi i/2}I_{\nu}\left(xe^{-3\pi i/4}\right),\]