Past exam of the mathematics course of the University of Cambridge / 2006 / ii / Paper 4 / 14E / a

Choose $F(z)=(z+1{)}^{1/3}{z}^{1/3}(z-1{)}^{1/3}$, each factor initially defined by polar argument in $(-\pi ,\pi )$. The products of the boundary values agree on $(-\infty ,-1)$, so their apparent cuts there cancel; the product extends analytically to $\mathbb{C}\setminus [-1,1]$ and has $F(z)\sim z$ at infinity. In polar coordinates about the three branch points this choice is $\prod r_j^{1/3}\exp [i\sum {\theta }_j/3]$.

For $0<x<1$, its upper and lower bank values are $e^{\pm i\pi /3}[x(1-x^2){]}^{1/3}$. For $-1<x<0$ they are $e^{\pm 2i\pi /3}[(-x)(1-x^2){]}^{1/3}$. The upper-minus-lower jump on each half is therefore $2i\sin (\pi /3)$ times the positive real integrand. Deforming a positively oriented large circle onto the cut, with the upper bank traversed right to left, givesSmall circles about the branch points contribute zero in the limit. At infinity,so the large-circle integral is $-2\pi i/3$. HenceBoth the branch choice and the orientation are essential to the sign.