\[\sin\left(\nu\pi\right){H^{(1)}_{\nu}}\left(ze^{m\pi i}\right)=-\sin\left((m-1%
)\nu\pi\right){H^{(1)}_{\nu}}\left(z\right)-e^{-\nu\pi i}\sin\left(m\nu\pi%
\right){H^{(2)}_{\nu}}\left(z\right),\]