Past exam of the mathematical methods I course of the University of Oxford / 2018 / 1 / a / Solution

Put $w=\mathrm{artanh}z$. The hyperbolic tangent identity $z=(e^{2w}-1)/(e^{2w}+1)$ gives $e^{2w}=(1+z)/(1-z)$. Taking a compatible branch of the complex logarithm givesAs an identity for a single-valued inverse hyperbolic tangent, this requires consistent branch cuts; one cannot choose unrelated logarithm branches on the two sides. Near $z=0$ the branch with value zero is fixed by $\mathrm{artanh}0=0$. The full inverse relation also allows $w\mapsto w+i\pi k$, $k\in \mathbb{Z}$, and $z=\pm 1$ are singular points.