Past exam of the mathematics course of the University of Cambridge / 2018 / ib / Paper 3 / 4A / b / ii / Solution

For $1<x<2$ the specified value is $g(x)=i\sqrt{(2-x)(x^2-1)}$. Continue towards $0$ around the branch point $1$. Along the upper bank, the argument of $(z-2)(z^2-1)$ increases from $\pi$ to $2\pi$, so its square root reaches the negative real value; along the lower bank it decreases from $\pi$ to $0$, giving the positive value. Since the radicand at $0$ is $2$,