Past exam of the mathematics course of the University of Cambridge / 2018 / ib / Paper 3 / 4A (Complex Methods)

On $\mathbb{C}\setminus [-1,1]$, the chosen branch of a multivalued function is characterized by $f(z{)}^2=z^2-1$ and $f(x)>0$ for $x>1$; equivalently $f(z)/z\to 1$ as $z\to \infty$. Define $h(z)=-f(-z)$. Then $h(z{)}^2=z^2-1$ and $h(z)/z\to 1$ at infinity, so uniqueness of the normalized holomorphic square root gives $h=f$. Therefore $f(-z)=-f(z)$.

Near infinity, $g(z{)}^2=(z-2)(z^2-1)\sim z^3$, so $g(z)\sim z^{3/2}$. One positive circuit around infinity changes $z^{3/2}$ by the monodromy factor $e^{3\pi i}=-1$. Thus analytic continuation does not return to the original value, and $\infty$ is a branch point.

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 $2$,