Past exam of the mathematics course of the University of Cambridge / 2026 / ii / Paper 3 / 30D / a

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