= Solution
Put $t=u+iv$ and $\Phi(t)=\sinh t-t$. Since
$$
\Phi(u+iv)=\sinh u\cos v-u+i(\cosh u\sin v-v),
$$
a <method of steepest descent>[steepest-descent curve] through the <saddle point> $t=0$ lies in
$$
\boxed{\cosh u\sin v-v=0.}
$$
The saddle is cubic rather than simple:
$$
\Phi(t)=\frac{t^3}{6}+O(t^5).
$$
Thus the curves meet the origin in the three descent directions for which $t^3$ is negative real,
$$
\boxed{\arg t=\frac\pi3,\ \pi,\ \frac{5\pi}3.}
$$
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, $v_+(u)\sim\sqrt3u$, while for $u\to+\infty$ the defining equation gives $\sin v_+\sim2v_+e^{-u}$, so $v_+\to\pi$. Their three asymptotes are therefore
$$
\boxed{v=0\quad(u\to-\infty),\qquad v=\pm\pi\quad(u\to+\infty).}
$$
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