= Solution
Write
$$
\lambda=Le^{i\pi/4},
\qquad
L=|\lambda|.
$$
The saddle equation is unchanged, so
$$
\boxed{s_+=Le^{i\pi/4},
\qquad s_-=-Le^{i\pi/4}=Le^{5i\pi/4}}.
$$
The contour connected to the original positive-real integration path passes through $s_+$. Since $\Phi(s_+)=0$, its steepest-descent branch has
$$
\operatorname{Im}\Phi(s)=0.
$$
For $s=re^{i\psi}$ and $\lambda^2=iL^2$,
$$
\operatorname{Im}\Phi
=-r\sin\psi+\sqrt2L
-\frac{L^2}{r}\cos\psi.
$$
Thus the contour equation is
$$
\boxed{
\frac rL\sin\psi
+\frac Lr\cos\psi=\sqrt2}.
$$
It approaches the origin along the positive imaginary axis, passes through $r=L$, $\psi=\pi/4$, and tends to infinity along the positive real axis. Along both ends the real part of the phase tends to $-\infty$, so the contour rises from the origin nearly vertically, bends through the first-quadrant saddle, and leaves nearly horizontally. The other saddle lies on a different constant-imaginary-phase contour and does not connect to this deformation.
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