Past exam of the mathematics course of the University of Cambridge / 2026 / ia / Paper 1 / 3E (Analysis I)

The radius is $R=\sup \{r\ge 0:\sum a_n{z}^n\text{ converges whenever }|z|<r\}$; equivalently $R^{-1}=\mathrm{limsup}|a_n{|}^{1/n}$. The same definition applies over $\mathbb{R}$ and $\mathbb{C}$.

Here $a_{2n}=(-1{)}^n$ and $a_{2n+1}=0$, so $\sum a_n{z}^n=\sum (-z^2{)}^n=(1+z^2{)}^{-1}=((z-i)(z+i){)}^{-1}$ for $|z|<1$.

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 $\mathrm{limsup}|a_n{|}^{1/n}=1$, $R=1$.

No. For example ${\sum }_{n\ge 0}(-1{)}^n{x}^n=(1+x{)}^{-1}$ has radius $1$ and is bounded by $1$ on $(0,1)$.