 Solution

Put $t=u+iv$ and $\Phi (t)=\sinh t-t$. Sincea steepest-descent curve through the saddle point $t=0$ lies inThe saddle is cubic rather than simple:Thus the curves meet the origin in the three descent directions for which $t^3$ is negative real,One curve is the negative real axis $v=0$, $u<0$. The other two are conjugate curves $v=\pm v_{+}(u)$ with $u>0$ and $0<v_{+}(u)<\pi$. Near the origin, $v_{+}(u)\sim \sqrt{3}u$, while for $u\to +\infty$ the defining equation gives $\sin v_{+}\sim 2v_{+}{e}^{-u}$, so $v_{+}\to \pi$. Their three asymptotes are therefore