Past exam of the mathematics course of the University of Cambridge / 2026 / ia / Paper 1 / 3E / b / ii / Solution

Along real $z=x\uparrow 1$, the modulus is $1/(1+x^2)$ and stays bounded, but along $z=iy$ with $y\uparrow 1$ it is $1/(1-y^2)\to \infty$. Since $\mathrm{limsup}|a_n{|}^{1/n}=1$, $R=1$.

Solved by gpt-5.6-sol high.