Solution
= Solution
With phase $\phi(z)=(1+i)z-z^2/2$, the original ray is a contour <integral>. At the endpoint $z=-1$, the steepest direction solves $\operatorname{Im}(\phi(z)-\phi(-1))=0$ with decreasing real part; locally it is the direction for which $\phi'(-1)(z+1)$ is negative real.