\[\operatorname{ker}_{\nu}x+i\operatorname{kei}_{\nu}x=e^{-\nu\pi i/2}K_{\nu}%
\left(xe^{\pi i/4}\right)=\tfrac{1}{2}\pi i{H^{(1)}_{\nu}}\left(xe^{3\pi i/4}%
\right)=-\tfrac{1}{2}\pi ie^{-\nu\pi i}{H^{(2)}_{\nu}}\left(xe^{-\pi i/4}%
\right).\]